Integrating $\int_{0}^{\infty}\prod_{k=1}^{n}(kx-1)e^{-ax} {d}x,$ where $a$ and $n$ are parameters. I want to integrate:

$\int_{0}^{\infty}\prod_{k=1}^{n}(kx-1)e^{-ax} {d}x, \quad $where $a$ and $n$ are parameters. Can I have a simple expansion of $\prod_{k=1}^{n}(kx-1)$?

Or can I use other techniques to figure it out? I just want the result to be simple.
 A: I will assume that $a \in \mathbb{R}$ with $a>0$. Start by computing the product of the integrand. Following (13) of this source,
$$
\prod_{k=1}^n (kx-1) = (-1)^n\sum_{k=0}^n S(n+1, k+1) x^{n-k}\tag{1}\label{eq:prod},
$$
where $S(n,k)$ denotes the signed Stirling number of the first kind. Plugging \eqref{eq:prod} into the integral, see that
$$
\begin{aligned}
\int_0^\infty \prod_{k=1}^n (kx-1)e^{-ax} \mathrm{d}x &= (-1)^n\sum_{k=0}^nS(n+1,k+1)\int^\infty_0 x^{n-k}e^{-ax} \mathrm{d}x \\
&= (-1)^n\sum_{k=0}^nS(n+1,k+1) \left[-\frac{\Gamma(n-k+1, ax)}{a^{n-k+1}}\right]^\infty_{x=0} \\
&= \left.(-1)^{n+1}\sum_{k=0}^nS(n+1, k+1)\frac{(n-k)!}{a^{n-k+1}}e^{-ax}\sum_{l=0}^{n-k}\frac{(ax)^l}{l!}\right|^\infty_{x=0},
\end{aligned}
$$
where $\Gamma(m, x)$ is the upper incomplete gamma function (see (2) for its expression with $m$ as an integer). Simplifying the last equality,
$$
\boxed{\int_0^\infty \prod_{k=1}^n (kx-1)e^{-ax} \mathrm{d}x = (-1)^n \sum_{k=0}^n S(n+1, k+1) \frac{(n-k)!}{a^{n-k+1}}.}
$$
