Proving the result $$(x+n)^n-\binom{n}{1}(x+n-1)^{n}+\binom{n}{2}(x+n-2)^{n}\cdots(-1)^nx^n=n!$$
Try: $$\bigg(x^n+\binom{n}{1}x^{n-1}n+\binom{n}{2}x^{n-2}n^2+\cdots n^n\bigg)-\binom{n}{1}\bigg(x^n+\binom{n}{1}x^{n-1}(n-1)\cdots\bigg)+\cdots +(-1)^n\binom{n}{n}x^n$$
Could some help me to solve it, thanks
It's convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write e.g. $$n![z^n]e^{kz}=k^n$$
We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n}{k}(x+n-k)^n(-1)^k}\\ &=\sum_{k=0}^n\binom{n}{k}(x+k)^n(-1)^{n-k}\tag{1}\\ &=\sum_{k=0}^n\binom{n}{k}n![z^n]e^{(x+k)z}(-1)^{n-k}\tag{2}\\ &=n![z^n]e^{xz}\sum_{k=0}^n\binom{n}{k}\left(e^z\right)^k(-1)^{n-k}\tag{3}\\ &=n![z^n]e^{xz}\left(e^z-1\right)^n\tag{4}\\ &=n![z^n]e^{xz}\left(z+\frac{z^2}{2!}+\cdots\right)^n\tag{5}\\ &\color{blue}{=n!} \end{align*} and the claim follows.
Comment:
In (1) we change the order of summation by setting $k \rightarrow n-k$ and use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
In (2) we apply the coefficient of operator.
In (3) we use the linearity of the coefficient of operator and do some rearrangements in order to apply the binomial theorem.
In (4) we apply the binomial theorem.
In (5) we do the series expansion of $e^{z}$ and see the smallest exponent of $z$ in $(e^z-1)^n$ is $\color{blue}{n}$. So, we will only use the constant $\color{blue}{1}$ in the series expansion of $e^{xz}=\color{blue}{1}+xz+\frac{(xz)^2}{2}+\cdots$ in order to obtain the coefficient of $z^n$, resulting in $$[z^n]e^{xz}\left(e^z-1\right)^n=1$$
I apply a one-to-one continuation of Jack D'Aurizio's answer to your (-;) question in here.
Let $p(x)$ be a polynomial with degree $d\geq 1$, and let $\delta$ be the (difference) operator mapping $p(x)$ into $p(x)-p(x-1)$. Then $(\delta p)(x)$ is a polynomial with degree $d-1$. We have that
$$(\delta^n p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k} p(x-k) \tag{1}$$
The sum you ask for is the RHS of $(1)$ in the case $p(x)=(x+n)^{n}$. So the original polynomial had degree $d=n$, hence $(1)$ has degree $n-n=0$ in $(x+n)$ which is to say that the result is the sum of all terms with power zero of $(x+n)$. This is given as
$$ \sum_{k=0}^{n}(-1)^k \binom{n}{k} (-k)^n= (-1)^n \sum_{k=0}^{n}\binom{n}{k} (-1)^k \frac{\partial^n}{(\partial a)^n}\exp (ak) |_{a=0} \\ = (-1)^n \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} \sum_{k=0}^{n}\binom{n}{k} (-e^a)^k = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} (e^a-1)^n \\ = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} a^n(1 +a/2 + \dots )^n = \lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n} a^n = n! $$
where in the last line, after expansion of the exponential, the fact was used that only terms with $a^n$ survive the operator $\lim_{(a \to 0)}\frac{\partial^n}{(\partial a)^n}$.
This proves the claim. $\qquad \Box$
