Prove that $\sum_{k=0}^n \frac{(-1)^{(n+k)}\sum_{r=0}^n a_rk^r}{k!(n-k)!}=a_n\quad \forall n \in \mathbb{N_0} $ Can Someone prove that the following equation is true? I tried to prove it using induction but I didn't get very far.
$$\sum_{k=0}^n \frac{(-1)^{(n+k)}\sum_{r=0}^n a_rk^r}{k!(n-k)!}=a_n\quad \forall n \in \mathbb{N_0} $$
Where $\sum_{r=0}^n a_rk^r$ is a polynomial of degree $n$ and $a_n$ is the coefficient of the highest power of that polynomial. I know that for $k=0$ and $r=0$ you get $0^0$ which ofcourse is undefined. But can we just say for the sake of simplicity that $0^0=1$?
$$\sum_{k=0}^n \frac{(-1)^{(n+k)}p(k)}{k!(n-k)!}=\frac{\partial^n p}{\partial k^nn!} \quad \forall n \in \mathbb{N_0} $$
Where $p(k)$ is a polynomial of degree $n$ should be an equivalent statement. This is my first question here by the way. If I made some mistakes, please let me know.

$\begin{array}\\
s_n
&=\sum_{k=0}^n \frac{(-1)^{(n+k)}}{k!(n-k)!}\sum_{r=0}^n a_rk^r\\
&=\frac{(-1)^n}{n!}\sum_{k=0}^n \frac{(-1)^{k}n!}{k!(n-k)!}\sum_{r=0}^n a_rk^r\\
&=\frac{(-1)^n}{n!}\sum_{k=0}^n (-1)^{k}\binom{n}{k}\sum_{r=0}^n a_rk^r\\
&=\frac{(-1)^n}{n!}\sum_{r=0}^n\sum_{k=0}^n (-1)^{k}\binom{n}{k} a_rk^r\\
&=\frac{(-1)^n}{n!}\sum_{r=0}^na_r\sum_{k=0}^n (-1)^{k}\binom{n}{k} k^r\\
&=\frac{(-1)^n}{n!}(-1)^na_nn!\\
&=a_n\\
\end{array}
$
This is because
$\sum_{k=0}^n (-1)^{k}\binom{n}{k} k^r
$
is the $n$-th difference
of a $r$-th degree polynomial
which is zero for
$r < n$
and $n!$ times the coefficient of
$x^n$
for $r = n$.
