Combinatoric proof for $\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$ ($n\geqslant5$) I'm trying to prove the following:
For every $n \ge 5$:
$$\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$$
I've tried cancelling one $(n-k)$, and got this:
$$n\sum_{k=0}^{n-1}{n-1\choose k}\left(-1\right)^k\left(n-k\right)^3 = 0$$
I've also tried expressing the first formula as such:
$$\sum_{k=0}^na_kb_k$$
Where $a_k = {n \choose k}\left(-1\right)^k$ and $b_k = \left(n-k\right)^4 = \sum_{j=0}^4{4\choose j}n^j\left(-k\right)^{4-j}$
It's easy to see that $\sum_{k=0}^n a_k = \left(1-1\right)^n = 0$ by the binomial theorem.
But I'm lost as to why this work only for n>=5. What am I missing?

It will be useful to know:
$\mathbf{ Theorem.}$ Let    $p(x)= a_0+a_1x+\cdots +a_nx^n$ be  $\mathit{any}$   polynomial in $\mathbb{C}[x]$ (of degree $\leq n$), then
$$ \sum_{k} {n\choose k}(-1)^k p(k)=(-1)^n n! a_n.$$
So in particular when $p$ has degree<n, then such sums are $0$.
Proof: see Graham, Knuth, Patashnik, Concrete Mathematics, Addison Wesley, 1989 , formula (5.42).
(As was suggested  above the proof is by using difference operators. ) $\Box$
Concerning the original question, since   $x\mapsto (n-x)^4$ is of degree 4 in $x$ but $n\geq 5 $ we get 0.
