I would like to compute the following summation of series: $$F(k,a)=\sum_{m=-1}^{k}(-1)^m\frac{k!}{(k-m)!}a^{k-m},$$ where $a$ is a known constant, $k$ and $m$ are integers. $x!$ indicates the factorial value of $x$.
WolframAlpha gave me a result: $$F(k,a)=(-1)^ke^{-a}\Gamma(k+1,-a)-\frac{a^{k+1}}{k+1},$$ considering k is an integer: $$\Gamma(k+1,-a)=k!e^a\cdot e_k(-a),$$ then $$F(k,a)=(-1)^kk!e_k(-a)-\frac{a^{k+1}}{k+1},$$ here $e_k(x)$ is the exponentional sum function: $e_k(x)=\sum_{n=0}^{k}\frac{x^n}{n!}.$
I have checked that this result is correct for $k=0$. When assuming that it also holds for $k=N$, how to prove the correctness for $k=N+1$ using the constructive proof method?
Thanks very much indeed!
Changing the summation index to $\,n=k-m\,$ gives the following, equivalent to the posted form:
$$ \begin{align} F(k,a) &= \sum_{n=0}^{k+1}\,(-1)^{k-n}\,\frac{k!}{n!}\,a^{n} \\[5px] &= (-1)^k\, k!\, \sum_{n=0}^{k+1}\,\frac{(-1)^{-n}a^{n}}{n!} \\[5px] &= (-1)^k\, k!\, \sum_{n=0}^{k+1}\, \frac{(-a)^{n}}{n!} \\[5px] &= (-1)^k \,k!\, e_{k+1}(-a) \end{align} $$