Summation of series involving binomial coefficients and polynomial of degree at most n-1 Prove that if $p$ is a polynomial of degree at most $n-1$, then 
$$\sum_{k=0}^n \binom{n}{k}(-1)^k p(k) = 0.$$
This was probably asked before but I couldn't find it anywhere. Can anyone please help me out here? Looks like the proof is extensive.
 A: Let $(\Delta f)(x) = f(x) - f(x+1)$. Then your expression is $(\Delta^np)(0)$ (so apply $\Delta$ $n$ times.) Now note that $\Delta$ on a constant function is $0$ and that it lowers the degree of a non-constant polynomial by one.
A: Consider the binomial expansion $(e^p -1)^n = \sum_{k=0}^n \binom{n}{k} (-1)^k e^{pk}$. Then note that
\begin{align}
\frac{\partial^m}{\partial p^m}(e^p-1)^n \bigg |_{p=0} & = \begin{cases} 0 & \text{if } m < n \\ n! & \text{if } m = n \end{cases}.
\end{align}
The important thing from this relation is that if $m \ge n$ the above expression is nonzero. Then we see $\sum_{k=0}^n \binom{n}{k} (-1)^k k^m = 0$ for $m < n$.
From here it is clear that a linear combination $p(k) = \sum_{m=0}^{n-1} a_m k^m$ will work.
A: As a slight variation of WimC answer: you can also apply induction, using $\binom{n}{k} = \binom{n-1}{k-1}-\binom{n-1}{k}$. You get, for $n-1$, a similar equation as the original with the difference polynomial $p_{n-1}(k)=p_n(k+1)-p_n(k)$. You only need to know/prove that the first difference of a polynomial decreases its degree by one.
