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.

$$\sum_{k=0}^n \frac{(-1)^{(n+k)}p(k)}{k!(n-k)!}= a_n$$
$$n!\sum_{k=0}^n \frac{(-1)^{(n+k)}p(k)}{k!(n-k)!}=n!\cdot a_n$$
$$\sum_{k=0}^n \frac{n!}{k!(n-k)!}(-1)^{(n+k)}p(k)=n!\cdot a_n$$
$$\sum_{k=0}^n (-1)^{n+k}\binom{n}{k}p(k)=n!\cdot a_n $$
Above identity is known as the Nörlund–Rice integral.
