TITLE: Are there any identities for alternating binomial sums of the form $\sum_{k=0}^{n} (-1)^{k}k^{p}{n \choose k} $?
QUESTION [10 upvotes]: In equations (20) - (25) of Mathworld's article 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 $D<n$.

REPLY [6 votes]: Up to the factor $(-1)^n$, the uncut sum is
$$s_{p,n}:=\sum_{k=0}^n(-1)^{n-k}\, k^p\,\binom nk.$$
As noted in the comment by Richard Stanley,
$$s_{p,n}=(\Delta^n f_p)(0),$$
where $f_p(x):=x^p$ and $(\Delta f)(x):=f(x+1)-f(x)$. Here and in what follows, $x$ denotes any real number.
It is easy to check by induction on $n$ that for any smooth enough function $f$ we have
$$(\Delta^n f)(x)=Ef^{(n)}(x+S_n),$$
where $f^{(n)}$ is the $n$th derivative of $f$, $S_n:=U_1+\cdots+U_n$, and $U_1,\dots,U_n$ are independent random variables uniformly distributed on the interval $[0,1]$. So,
$$s_{p,n}=n!\binom pn ES_n^{p-n} \tag{1}$$
for $p=0,1,\dots$ and $n=0,1,\dots$. In particular, it follows that $s_{p,n}=0$ for $n=p+1,p+2,\dots$, as noted in the answer by Carlo Beenakker.
In fact, (1) holds for all real $p\ge n$ (and $n=0,1,\dots$), and then, obviously,
$$0<s_{p,n}\le n!\binom pn n^{p-n}. \tag{2}$$
If $p-n\ge1$, then, in view of Jensen's inequality, the lower bound $0$ on $s_{p,n}$ in (2) can be greatly improved, to $$b_{p,n}:=n!\binom pn \Big(\frac n2\Big)^{p-n}.$$
Moreover, by the law of large numbers, $S_n/n\to1/2$ in probability (say). Also, $0\le S_n/n\le1$. So, by dominated convergence, from (1) we immediately get the following asymptotics: if $n\to\infty$ and $p-n\to a$ for some real $a>0$, then
$$s_{p,n}\sim b_{p,n}.$$