Suggestions for proving $ \displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)=p^{(n)}(0)$ The polynomial $p$ has degrees less or equal to n
and I'm trying to prove
\begin{equation}
 \displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)=p^{(n)}(0)
\end{equation}
$p^{(n)}(0)=a_n\cdot n!$
The followings are my attempts
First is just to expand $\displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)$, but I didn't find any way of solving this.
Second I started to consider about using induction. Base case degree p=1 is true and suppose $\displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)=p^{(n)}(0)$
Prove
\begin{equation}
\displaystyle \sum^{n+1}_{k=0} \binom{n+1}{k}(-1)^{n+1-k}p(k)=p^{(n+1)}(0)
\end{equation}
Then I tried to just directly expand this formula, but it seemingly doesn't work and it requires a lot of computation
(as so far, the most hopeful way)
Then I use the fact $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$ to split the formula
My attempt was:
\begin{equation*}
\displaystyle  \sum^{n+1}_{k=0}[\binom{n}{k}+\binom{n}{k-1}](-1)^{n+1-k}p(k) 
\end{equation*}
and get
\begin{equation*}
\displaystyle  \sum^{n+1}_{k=0}\binom{n}{k}(-1)^{n+1-k}p(k) + \sum^{n+1}_{k=0}\binom{n}{k-1}(-1)^{n+1-k}p(k) 
\end{equation*}
For the left part, since when k=n+1, $\binom{n}{n+1}=0$,
\begin{equation*}
\displaystyle  \sum^{n+1}_{k=0}\binom{n}{k}(-1)^{n+1-k}p(k)=\displaystyle  \sum^{n}_{k=0}\binom{n}{k}(-1)^{n+1-k}p(k)=(-1)\sum^{n}_{k=0}\binom{n}{k}(-1)^{n-k}p(k)=-p^{(n)}(0)
\end{equation*}
However, For the right part:when k=0,$\binom{n}{−1}=0$. Also, let k=j+1 thus it can be
\begin{equation}
\sum^{n+1}_{k=0}\binom{n}{k-1}(-1)^{n+1-k}p(k)= \sum^{n+1}_{k=1}\binom{n}{k-1}(-1)^{n+1-k}p(k)=\sum^{n}_{j=0}\binom{n}{j}(-1)^{n-j}p(j+1)
\end{equation}
Then I totally get stuck, since I don't know how to deal with this $p(j+1)$, and I can't come up with more ideas. Any help, comments, or suggestions are appreciated!

Based on the idea from pirahahindu1999, my trial:
regard the formula $\displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)$ as a map, and all polynomials are vectors.
Thus we have $F(p(x))=\displaystyle \sum^n_{k=0} \binom{n}{k}(-1)^{n-k}p(k)$
It's true that the map $F$ is a well-defined and linear map.
Then $F$ :$ P_n(x) \to \mathbb R$
It's clear that $P_n(x)$ has a basis $\{1,x,x(x-1),x(x-1)(x-2),...,x(x-1)...1\}$ and suppose $1=e_0,x=e_1,x(x-1)=e_2...$
Then it's true that $F$ depends on the degree of the basis vectors, and the fact is that $F_3(e_3)=3!,F_3(e_2)=F_3(e_1)=F_3(e_0)=0$ Similarly for $F_n$
Since any polynomial can be expressed as the linear combination of basis vectors, it's true to have
\begin{equation}
p(x)=a_0e_0+...+a_me_m
\end{equation}
When m=n, $F_n(p(x))=F(a_ne_n)=a_n·n!=p^{(n)}(0)$
When m<n, $F_n(p(x))=0$
qed
 A: The identity you are trying to prove is linear in the polynomial $p$ (where $p$ ranges over the elements of the vector space $\mathbb{R}_n[x]$ of polynomials of degree at most $n$), it's also an easy fact that $\mathbb{R}_n[x]$ has a basis $\{1,\frac{x}{1!},\frac{x(x-1)}{2!},\frac{x(x-1)(x-2)}{3!},....\} $. Thus it suffices to verfiy your identity for members of this basis. The binomial theorem will be helpful
A: Try proving this first when $p(k)= k^p$ is a monomial pure  power with $p<m$.
Another tip: You can also consider the the binomial expansion of $ (1- e^{ax})^m$ and obtain various combinatorial identities by differentiating this with respect to $a$ and evaluating at $a=0$.
A: We’ll use a combinatorial argument.
If $p(x)=x^m,$ this sum is the inclusion-exclusion count for the number of onto maps from a set of $m$ elements to a set on $n$ elements (see below.) When $n=m,$ this is $n!,$ and when $0\leq m<n,$ the number of elements is zero.
Then use the fact that the value is linear in $p.$

Inclusion-exclusion Argument
If $X=\{1,2,\dots,m\}$ and $Y=\{1,2,\dots,n\},$ let $A$ be the set of all functions $X\to Y.$ For each $i\in Y,$ let $A_i$ be the set of functions $f:X\to Y$ such that $f(x)\neq i$ for all $x\in X.$
The set of onto maps $X\to Y$ can be written $A\setminus(A_1\cup\cdots\cup A_n).$ Then inclusion-exclusion gives us:
$$\left|A\setminus(A_1\cup\cdots \cup A_n)\right|=\sum_{i=0}^{n}(-1)^i\binom ni (n-i)^m.$$
Substituting $k=n-i,$ or $i=n-k,$ you get your sum (after remembering $\binom n{n-k}=\binom nk.$)

These numbers are $n!S(m,n)$ where $S(m,n)$ are the Stirling numbers of the second kind.
