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.
 A: It is rather common in discrete math to define $0^0 = 1$. For example it is known that $\sum_{k = 0}^n (-1)^k\binom{n}{k} = (1 - 1)^n = 0^n$ that should be true for $n = 0$ too: $\binom{0}{0} = 1$. More arguments you can find here.

Some textbooks leave the quantity $0^0$ undefined, because the functions $0^x$ and $x^0$ have different limiting values when $x$ decreases to $0$. But this is a mistake. We must define
  $$x^0=1 \text{ for all }x,$$
  if the binomial theorem is to be valid when $x=0$, $y=0$ , and/or $x=-y$. The theorem is too important to be arbitrarily restricted! By contrast, the function $0^x$ is quite unimportant.

from Concrete Mathematics, p. 162, R. Graham, D. Knuth, O. Patashnik, Addison-Wesley, 1988
However note that there are some cases in discrete math when $0^0$ should be defined some other way, e. g., $0^0 = 0$ or $0^0 = 2$.
A: Euler proved that
$$ \sum_{k=0}^{n} (-1)^k \binom{n}{k} k^r = 0 $$
for integers $r<n$ (E368, §§ 25–26, see also Gould's article). One can prove this by applying the operator $x \frac{d}{dx}$ $r$ times to both sides of the identity
$$ (1-x)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^k $$
and evaluating at $x=1$. Applying the operator $n$ times instead gives
$$ \sum_{k=0}^{n} (-1)^k \binom{n}{k} k^n = (-1)^n n!, $$
and the result follows from here by multiplying both sides of this by $(-1)^n/n!$ and using linearity.
A: $\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 $r = n$.
A: $$\sum_{k=0}^n \frac{(-1)^{(n+k)}p(k)}{k!(n-k)!}= a_n$$
$$n!\sum_{k=0}^n \frac{(-1)^{(n+k)}p(k)}{k!(n-k)!}=n!\cdot a_n$$
$$\sum_{k=0}^n \frac{n!}{k!(n-k)!}(-1)^{(n+k)}p(k)=n!\cdot a_n$$
$$\sum_{k=0}^n (-1)^{n+k}\binom{n}{k}p(k)=n!\cdot a_n $$
Above identity is known as the Nörlund–Rice integral.
