I want to prove that the function $$g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))},\,x >0$$ is increasing in $x$ for all $n$, where $ S_n(x)= \sum_{m=0}^{n}\frac{x^m}{m!}$. Differentiating gives something messy that I have not been able to prove it is non-negative. I have also been trying to find an appropriate $h(x)$ increasing and proving that $ h(g(x))$ is increasing $\ln(\cdot)$ seems a good candidate.
By plotting I am almost convinced the statement is true.
Note: Here is a first step towards a complete answer. The point I like to address is the discrete log-concavity of the function $S_n$ for fixed $x>0$. This means $S_n$ fulfils
\begin{align*} S_{n-1}^2(x)\geq S_n(x)S_{n-2}(x)\qquad\qquad x>0, n\geq 2\tag{1} \end{align*}
A corresponding example can be found e.g. in the slides about Log-convexity and Log-concavity by Dmitry Karp. You might have a look at the section about Bessel functions where the inequality (1) for these functions is stated and some sharper results are presented afterwards.
Please note that in order to show $g(x)$ is increasing an even sharper inequality than (1) has to be proved.
We start as @Arashium did. Using the differential operator $D_x$ we obtain \begin{align*} D_xS_n(x)=D_x\sum_{m=1}^{n}\frac{x^{m-1}}{(m-1)!}=\sum_{m=0}^{n-1}\frac{x^m}{m!}=S_{n-1}(x) \end{align*} and by omitting the argument $x$ we get \begin{align*} D_x\ln(S_n)&=\frac{1}{S_n}D_x(S_n)=\frac{S_{n-1}}{S_n}\\ \end{align*} and \begin{align*} D_xg&=D_x\frac{\ln(S_n)}{\ln(S_{n-1})} =\frac{\ln(S_{n-1})D_x\ln(S_n)-\ln(S_n)D_x\ln(S_{n-1})}{\ln^2(S_{n-1})}\\ &=\frac{1}{\ln^2(S_{n-1})}\left(\frac{S_{n-1}}{S_n}\ln(S_{n-1})-\frac{S_{n-2}}{S_{n-1}}\ln(S_n)\right)\tag{2} \end{align*}
In order to show that the function $g$ is increasing we need $D_xg\geq 0$.
Therefore the following is to prove according to (2) \begin{align*} S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}\tag{3} \end{align*}
Here's the proof of the weaker inequality (1) showing that $S_n(x)$ is discrete log-concave for fixed $x>0$.
\begin{align*} S_{n-1}(x)^2&-S_{n-2}(x)S_n(x)=\\ &=\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}\right)^2-S_{n-2}(x)\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n!}\right)\\ &=S_{n-2}(x)\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^{n}}{n!}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}S_{n-2}(x)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}\right)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=0}^{n-2}\frac{x^{m+1}}{m!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=1}^{n-1}\frac{x^{m}}{(m-1)!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}-\frac{x^{n-1}}{n(n-2)!}\right) +\frac{x^{2n-2}}{(n-1)!^2}\\ &=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}\right)\\ &\qquad\qquad+\frac{x^{2n-2}}{(n-1)!(n-2)!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{4}\\ &>0\\ \end{align*}
From the line (4) it's obvious that the inequality is valid and we may therefore conclude that $S_n$ is discrete log-concave.
The challenge is of course to sharpen the inequality (4) in order to obtain (3).
Note: Since the function $\ln$ is monotonically increasing we obtain due to the log-concavity of $S_n$
\begin{align*} \ln\left(S_{n-1}^2(x)\right)&\geq\ln\left(S_n(x)S_{n-2}(x)\right)\\ 2\ln\left(S_{n-1}(x)\right)&\geq\ln\left(S_n(x)\right)+\ln\left(S_{n-2}(x)\right)\\ \end{align*}
which may of some use for further calculations.
I can reduce the problem to a purely polynomial identity, without any logs in it. As explained by both @MarkusScheuer and @Arashium, the inequality to be shown is equivalent to $S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}$. To isolate one of the logs, let us put $\phi_n=\frac{S_{n-1}^2}{S_{n-2}S_n}\ln(S_{n-1})-\ln(S_n)$. The goal is then to show that $\phi_n$ is nonnegative. Since $\phi_n(0)=0$, it will suffice to show that $\phi'_n$ is nonnegative. If we set $F_n=\frac{S_{n-1}^2}{S_{n-2}S_n}$, then
$$\phi'_n=F'_n\ln(S_{n-1})+F_n\frac{S_{n-2}}{S_{n-1}}-\frac{S_{n-1}}{S_{n}}= F'_n\ln(S_{n-1}) \tag{1} $$
Since $\ln(S_{n-1}) \geq 0$, it suffices to show that $F'_n$ is nonnegative. A little computation shows that
$$ F'_n=\frac{S_{n-1}}{(S_{n-2}S_n)^2}G_n, \ \text{with} \ G_n=2S_nS_{n-2}^2-S_{n-1}(S_{n-3}S_n+S_{n-2}S_{n-1}) $$
It will suffice to show that the rescaled polynomial $H_n=\frac{n!(n-1)!(n-2)!}{x^{n-2}}G_n$ satisfies
$$ H_n=\sum_{0 \leq i \leq j \leq n-2} (n-j)!(n-j-1)!(j-i)! \binom{n-2}{j}\binom{j}{i}\binom{2n+1-i-j}{j-i}x^{i+j} \tag{2} $$
I have checked that (2) is true for any $n\leq 40$ with the help of a computer, but failed to find a proof so far. I created a separate question for the proof of (2).
I'M SORRY, THIS ANSWER WAS BROKEN. In fact, $(\dagger)$ does not imply $(\ast)$. I'll leave it up for now, until I've figured out what best to do with it.
The point of this answer is to provide a reduction to a different polynomial inequality. Although I can't prove this inequality either, I am more optimistic about it, because it is quadratic in the $S_n$, instead of cubic like Ewan's, and because I can show it is logically equivalent to Ewan's inequality.
Let's recall that Ewan's inequality is $$-S_{n-1}^2 S_{n-2} + 2 S_n S_{n-2}^2 - S_{n} S_{n-1} S_{n-3} \geq 0 \quad (\ast).$$ The point of this answer is to show that $(\ast)$ is logically equivalent to $$\frac{n-1}{n} S_{n-1}^2 \leq S_n S_{n-2}. (\dagger)$$ For $2 \leq n \leq 30$, I've checked the stronger statement that the coefficients of $S_n S_{n-2} - \frac{n-1}{n} S_{n-1}^2$ are positive.
Proof that $(\ast)$ implies $(\dagger)$: Dividing by $S_n S_{n-1} S_{n-2}$, $(\ast)$ is $$0 \leq - \frac{S_{n-1}}{S_n} + 2 \frac{S_{n-2}}{S_{n-1}} - \frac{S_{n-3}}{S_{n-2}} = \frac{d}{dx} \left( \log \frac{S_{n-1}^2}{S_{n} S_{n-2}} \right)$$ So $(\ast)$ is equivalent to the claim that $\frac{S_{n-1}^2}{S_{n} S_{n-2}}$ is increasing. We note that $\lim_{x \to \infty} \frac{S_{n-1}^2}{S_{n} S_{n-2}} = \frac{n}{n-1}$ so, if $(\ast)$ holds, then $\frac{S_{n-1}^2}{S_{n} S_{n-2}} \leq \frac{n}{n-1}$ for all $x$, which rearranges to $(\dagger)$. $\square$
Incidentally, we can also use this to give another proof that $(\ast)$ is equivalent to the original inequality: $\frac{S_{n-1}^2}{S_{n} S_{n-2}} = \frac{S_{n-1}/S_{n}}{S_{n-2}/S_{n-1}}$. If $f(x)/g(x)$ is increasing, then so is $\int_0^y f(x) dx/\int_0^y g(x) dx$ (see, for example, [here][1]). So $\frac{\int S_{n-1}/S_{n}}{\int S_{n-2}/S_{n-1}} = \frac{\log S_{n}}{\log S_{n-1}}$ is increasing.
Proof that $(\dagger)$ implies $(\ast)$. Starting with $(\dagger)$, multiply by $S_n S_{n-2} x^{2n-2}/(n! (n-1)!)$ to get
$$S_n S_{n-1}^2 S_{n-2} \frac{x^n x^{n-2}}{n! (n-2)!} \geq \frac{(x^{n-1})^2}{(n-1)!^2} S_n^2 S_{n-2}^2.$$
I miscopied the sign at this step; the inequality in the middle should point the other way. I've deleted the steps after this, but the idea was to get to do some algebra, get to an AM-GM, and then do some more algebra. Since the AM-GM is the only non-reversible step, this means we can't prove $(\ast)$ by this route.
In other words, $$(1/2) (x^n/n! S_{n-1} S_{n-2} + x^{n-2}/(n-2)! S_n S_{n-1}) \geq S_n (x^{n-1}/(n-1)!) S_{n-2} \quad (\ast \ast)$$ is presumably true, since it is logically equivalent to $(\ast)$. But $$\sqrt{x^n/n! S_{n-1} S_{n-2} \cdot x^{n-2}/(n-2)! S_n S_{n-1}} \geq S_n (x^{n-1}/(n-1)!) S_{n-2}$$ is false already for $n=2$. So any attempt to prove $(\ast \ast)$ must do something more gentle than AM-GM.
I don't have a solution here but just write down what I have reached till now.
I cannot solve it now since I am in rush to leave. But it may help some others to continue.
To show $\mathfrak{g}(x)$ is increasing: $$\mathfrak{g}(x)=\frac{\ln f}{\ln g}$$
$$\mathfrak{g}'(x)\ge0$$
$$\mathfrak{g}'(x)=\frac{\frac{f'}{f}\ln g -\frac{g'}{g}\ln f}{(\ln g)^2}\ge 0$$
or
$$\frac{f'}{f}\ge \frac{g'}{g}\frac{\ln f}{\ln g}$$
here $$f(x)=S_{n}(x)$$ $$g(x)=S_{n-1}(x)$$ $$S'_{k}(x)=S_{k-1}(x)$$
So we need to show
$$\ln (S_{n-1}) S^2_{n-1} \ge S_{n}S_{n-2} \ln (S_n)$$
$$S^2_{n-1}=S^2_{n-2}+(\frac{x^{n-1}}{(n-1)!})^2+2\frac{x^{n-1}}{(n-1)!} S_{n-2}$$
$$S_{n}S_{n-2}=S^2_{n-2}+S_{n-2}(\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!})$$
The rest of the proof seems to be more complicated than the question. I am very glad if somebody can continue it.
I am thinking about inductive reasoning too.
I can provide half the combinatorial argument that Ewan is asking for. In Ewan's notation, $$G_n = - (S_n-x^n/n!) S_{n-1} S_{n-2} + 2 S_n (S_{n-1} - x^{n-1}/(n-1)!) S_{n-2} - S_n S_{n-1} (S_{n-2} - x^{n-2}/(n-2)!)$$ $$=\frac{x^n}{n!} S_{n-1} S_{n-2} - 2 \frac{x^{n-1}}{(n-1)!} S_n S_{n-2} + \frac{x^{n-2}}{(n-2)!} S_n S_{n-1}.$$ Note that this makes it clear why $G_n$ has terms between $x^{n-2}$ and $x^{3n-3}$ (and a little more work shows that the coefficient of $x^{3n-4}$ is $0$.)
I can deal with $x^m$ for $n \leq m \leq 2n-3$. Set $m=n+r$, so $0 \leq r \leq n-3$. (It is also easy to handle $r=-1$ and $r=-2$, but I don't want to write out the details.) We want to understand $$\frac{[x^r]( S_{n-1} S_{n-2})}{n!} - 2 \frac{[x^{r+1}] (S_{n} S_{n-2})}{(n-1)!} + \frac{[x^{r+2}] (S_{n} S_{n-1})}{(n-2)!} \quad (\ast)$$ where $x^q f(x)$ denotes the coefficient of $x^q$ in $f(x)$. Now, $S_n(x)$ is the first $n$ terms of the Taylor series for $e^x$. So, for $r$ in the stated range, $$(\ast) = \frac{[x^r]e^{2x}}{n!} - 2 \frac{[x^{r+1}] e^{2x}}{(n-1)!} + \frac{[x^{r+2}] e^{2x}}{(n-2)!} = \frac{2^r}{r! n!} - 2 \frac{2^{r+1}}{(r+1)! (n-1)!} + \frac{2^{r+2}}{(r+2)! (n-2)!}.$$
Factoring out $2^r/((r+2)! n!)$, what remains is $$(r+2)(r+1) - 4 (r+2)(n) + 4 n(n-1).$$
The region $(r+2)(r+1) - 4 (r+2)(n) + 4 n(n-1) \geq 0$ is the exterior of a parabola (shown in blue), the region $r \leq n-3$ is a half plane (shown in yellow). The line and the parabola meet at $(r,n) = (-2,1)$ and $(-1, 2)$ so there are no integer points where $r \leq n-3$ and $(r+2)(r+1) - 4 (r+2)(n) + 4 n(n-1) < 0$.
What I can't figure out is how to deal with the terms for $r>n-3$, where the $S_i S_j$ terms in $(\ast)$ no longer match $e^{2x}$ in the relevant degrees.
It's not hard to prove that $ dS_n/dx = S_{n-1} $, and since $ Sn(x) > 0 $ for all x > 0 and any natural n, the derivative of g(x) is positive for all x.
Instead of differentiating $g$ you may try with the definition of an increasing function. Given $x_1, x_2\in \mathbb{R}^{+}$ such that $x_1 < x_2$ then $g(x_1) < g(x_2)$.
You have that the $\ln$ is an increasing function and so is the $S_n$ for a fixed $n$, furthermore $S_n(x) > S_{n-1}(x),\ \forall x$ (and so its derivative, which is how the function $S$ varies).
Putting everything together, \begin{equation*} g(x_1) = \frac{\ln(S_n(x_1))}{\ln(S_{n-1}(x_1))} < \frac{\ln(S_n(x_2))}{\ln(S_{n-1}(x_2))} = g(x_2) \end{equation*}
and that's it.
