Relation between two polynomials I am stuck on the following exercise.
Let $P \in \mathbb{C}[X]$ be a polynomial of degree $n \in \mathbb{N}$, $n > 0$. We write :
$$ P(X) = \sum_{k=0}^{n} a_k X^k $$
with $a_0, \ldots, a_n \in \mathbb{C}^{n+1}$.
I would like to prove that :
$$ \sum_{k=0}^{n} (-1)^{n-k} \begin{pmatrix} n \\ k \end{pmatrix} P(k) = n! ~ a_n. $$
Could someone, please, suggest me ways/ideas to prove this equality ?
 A: The polynomials $\binom{X}{u}:u=0,\ldots,n$ are linearly independent over $\Bbb C$ and generates the set of all polynomials of degree $n$.
Thus there exists $b_u\in\Bbb C$ such that 
$$P(X)=\sum_{u=0}^nb_u\binom Xu$$
note that $n!a_n=b_n$.
Let $\mathrm D$ denote the formal derivative respect to $X$, then
\begin{align}
f_u(X)
&:=\sum_{k=0}^n\binom nk(-1)^{n-k}\binom ku X^{k-u}\\
&=\sum_{k=0}^n\binom nk(-1)^{n-k}\frac 1{u!}\mathrm D^u X^k\\
&=\frac 1{u!}\mathrm D^u\sum_{k=0}^n\binom nk(-1)^{n-k} X^k\\
&=\frac 1{u!}\mathrm D^u(X-1)^n\\
\end{align}
hence $f_u(1)=0$ for $u<n$ while $f_n(1)=1$.
Consequently,
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom nkP(k)
&=\sum_{u=0}^n b_uf_u(1)\\
&=b_n\\
&=a_nn!
\end{align}
A: Here is an idea. Both sides equal $(\Delta^nP)(X)$. Where $(\Delta P)(X)=P(X+1)-P(X)$. We claim that
$$
(\Delta^nP)(X)=n!a_{n}.
$$
To see this write
$$
 P(X) = \sum_{k=0}^{n} a_k X^k=  \sum_{k=0}^{n} a_k \sum_{m=0}^{k}S(k,m)X^{\underline{m}}\tag{1}
$$
where $S(k,m)$ is the stirling number of the second kind. and $X^{\underline{M}}$ is the falling factorial. Since $\Delta(X^{\underline m})=mX^{\underline{m-1}}$, $\Delta^{n}(X^{\underline {m}})=m^{\underline n}X^{\underline{m-n}}$. If $m<n$, then $\Delta^{n}(X^{\underline {m}})=0$. Hence taking $\Delta ^{n}$ of both sides of (1) yields that
$$
(\Delta^nP)(X)=n!S(n,n)a_{n}=n!a_{n}\tag{2}.
$$
Now write
$$
P(X)=b_{0}\binom{X}{0}+b_{1}\binom{X}{1}+\dotsb+b_{n}\binom{X}{n}\tag{3}
$$
Observe that
$$
\Delta\binom{X}{k}=\binom{x}{k-1}.
$$
Hence it follows that
$$
(\Delta^nP)(X)=b_{n}=\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} P(k). \tag{4}
$$
You can prove (4) by induction using the relationship.
$$
b_{n}=P(n)-b_{0}-b_{1}\binom{n}{1}-\dotsb-b_{n-1}\binom{n}{n-1}.
$$
Equations (2) and (4) yield the result.
