Combinatoric proof for $\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$ ($n\geqslant5$)
The expression counts the number of ways to partition a set of $4$ elements into $n$ distinguishable non-empty cells, which certainly must give zero for $n \geq 5$.
The result follows by inclusion exclusion. Alternately, see Stirling numbers of the second kind for a more general view.
$p(k)=k^4$ is a fourth degree polynomial. Let $\delta$ be the backward difference operator: $$ \delta f(x) = f(x)-f(x-1).$$ Since the degree of $\delta f$ is one less than the degree of $f$, for any $n\geq 5$ we have $\delta^n p(x) = 0$.
The idea of cancelling the factors of $n-k$ should work. $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^m &=n\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^{m-1}\\ &-n\sum_{k=0}^n(-1)^k\binom{n-1}{k-1}(n-k)^{m-1}\\ &=n\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}(n-k)^{m-1}\\ \end{align} $$ For $m\le n$, induction yields $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^m &=\frac{n!}{(n-m)!}\sum_{k=0}^{n-m}(-1)^k\binom{n-m}{k}\\ &=\frac{n!}{(n-m)!}(1-1)^{n-m}\\ &=\left\{ \begin{array}{} n!&\text{if }m=n\\ 0&\text{if }m\lt n \end{array} \right. \end{align} $$
A Second Approach
In this answer there are three proofs of $$ \begin{align} \sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k} &=\left\{\begin{array}{} 1&\text{if }n=k\\ 0&\text{otherwise} \end{array}\right.\\ &=[n=k] \end{align} $$ where $[\dots]$ are Iverson Brackets. Furthermore, $\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}}$ $$ \sum_{k=0}^m\binom{n}{k}\,\stirtwo{m}{k}k!=n^m $$ where $\stirtwo{m}{k}$ are Stirling Numbers of the Second Kind. Therefore, $$ \begin{align} \sum_{k=0}^n(-1)^k\binom{n}{k}(x-k)^m &=\sum_{k=0}^n\sum_{j=0}^m(-1)^{k-j}\binom{n}{k}\binom{m}{j}x^{m-j}k^j\\ &=\sum_{k=0}^n\sum_{j=0}^m\sum_{i=0}^j(-1)^{k-j}\binom{n}{k}\binom{m}{j}x^{m-j}\binom{k}{i}\stirtwo{j}{i}i!\\ &=\sum_{j=0}^m\sum_{i=0}^j(-1)^{n-j}\binom{m}{j}x^{m-j}\,[n=i]\,\stirtwo{j}{i}i!\\ &=\sum_{j=0}^m(-1)^{n-j}\binom{m}{j}x^{m-j}\stirtwo{j}{n}n! \end{align} $$ If $m\lt n$, then either $\binom{m}{j}=0$ or $\stirtwo{j}{n}=0$. If $m=n$, the only non-zero term is $j=m$.