The first column below is n (iterator), the second is the value, I'm trying to understand if there is a closed formula that I can use for obtaining the value given a fixed n
$$n=0: a$$ $$n=1: a(y-1) + 1$$ $$n=2: a(y^2-2y+2)-2$$ $$n=3: a(y^3-3y^2+6y-6)+6$$ $$n=4: a(y^4-4y^3+12y^2-24y+24)-24$$
Let $\beta$ be any number with $|\beta| < 1$. Notice
$$\begin{align} \sum_{n=0}^\infty \frac{\beta^n}{n!} \int_0^y x^n e^{x} dx &= \int_0^y e^{(1+\beta)x} dx = \frac{1}{1+\beta}\left(e^{(1+\beta)y} - 1\right)\\ &= \left[\sum_{p=0}^\infty(-1)^p\beta^p\right]\left[e^y\left(\sum_{q=0}^\infty \frac{\beta^q}{q!}y^q\right)-1\right] \end{align} $$ By comparing coefficients of $\beta^n$ on both sides, we get
$$\int_0^y x^n e^x dx = n! \left[ e^y \left( \sum_{q=0}^n \frac{(-1)^{n-q}}{q!}y^q\right) - (-1)^n\right]\tag{*1}$$
Ideas behind such a closed form
To shine some light why the closed from looks like this, let us re-derive the integral in a completely bogus manner. It is known that for any smooth function $f(x)$ over $\mathbb{R}$, we have $$\frac{d}{dx}\Big[ f(x) e^x \Big] = ( f(x) + f'(x) ) e^x\tag{*2}$$
If we view $\displaystyle\;D = \frac{d}{dx}\;$ as some sort of operator over the space of smooth functions. $(*2)$ can be reinterpreted as
$$ L^{-1} D L = D + 1$$ where $L$ is the operator of pointwise multiplication a function by $e^x$. In certain sense, integration is the inverse of differentiation. Expression of the form
$$e^{-y} \int^y e^{x} f(x) dx\tag{*3}$$
can be interpreted as applying the operator
$$L^{-1} D^{-1} L = ( L^{-1} D L )^{-1} = (1 + D)^{-1}$$ to the function $f$. If we close our eyes and expand
$$(1 + D)^{-1} = 1 - D + D^2 + D^3 - \ldots...$$ We get a formal expression for the integral
$$\int^y f(x) e^x dx = e^y ( f(y) - f'(y) + f''(y) - \ldots ) + \text{constant}\tag{*4}$$
Even though the logic is bogus, if we apply this to $f(x) = x^n$, we reproduce the formula $(*1)$.
In certain sense, $(*4)$ can be viewed as the result of repeatedly applying integration by parts to $f(x)$. When $f(x)$ is a polynomial, the chain of integration by parts terminated in finite steps and we get back another polynomial multiplying $e^x$. That's why $(*4)$ works when $f(x)$ is a polynomial and why $(*1)$ have such a form.
The coefficients are known as the falling factorial triangle, $T(N,K)=\frac{N!}{(N-K)!}$:
1
1, 1
1, 2, 2
1, 3, 6, 6
1, 4, 12, 24, 24
1, 5, 20, 60, 120, 120
1, 6, 30, 120, 360, 720, 720
1, 7, 42, 210, 840, 2520, 5040, 5040
1, 8, 56, 336, 1680, 6720, 20160, 40320, 40320