In equations (20) - (25) of Mathworld's [article](https://mathworld.wolfram.com/BinomialSums.html) on binomial sums, identities are given for sums of the form $$\sum\_{k=0}^{n} k^{p}{n \choose k}, $$ with $p \in \mathbb{Z}\_{\geq 0}$. I wonder whether identities also exist for the alternating counterparts: $$\sum\_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} .$$ Furthermore, I'm interested in results for the same sum that is “cut off”, i.e. when the summands go from $k=0$ to some $D<n$.



Are there any identities for alternating binomial sums of the form $\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} $?
A rewrite of formula (10) on [MathWorld](https://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html) (replacing the summation index $k-i\mapsto i$) gives the desired formula:
$$\sum\_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} =(-1)^n n! S\_2(p,n),$$
where $S\_2(p,n)$ is the [Stirling number of the second kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind) (the number of ways of partitioning a set of $p$ elements into $n$ non-empty subsets).  

It is remarkable that the alternating sum equals zero for $p<n$.


