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 $D<n$.
A rewrite of formula (10) on MathWorld (replacing the summation index $ki\mapsto i$) 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 (the number of ways of partitioning a set of $p$ elements into $n$ nonempty subsets).
It is remarkable that the alternating sum equals zero for $p<n$.

11$\begingroup$ It's not so remarkable that the alternating sum is zero for $p<n$. The sum is the $n$th difference at 0 of a polynomial of degree $p$. For any polynomial of degree $p$, the $p$th difference is constant and the $n$th difference for $n>p$ is 0. $\endgroup$ Aug 30 '20 at 21:48
For the cutoff version:
We can get a subtractionfree formula for the cutoff version, which should be sufficient to get asymptotics, by the same idea that gives a simple bijective proof of the identity that Carlo Beenakker mentioned. That is:
$k^p$ counts maps from a $p$element set $[p]$ to a $k$element set
Thus $\binom{n}{k} k^p$ counts pairs of a $k$element subset $S$ of an $n$elements set $[n]$ with a map from $[p]$ to $S$. In other words, it counts maps $f$ from $[p]$ to $[n]$ together with a $k$element subset $S$ of $[n]$ containing the image of $f$.
So $\sum_{k=0}^d (1)^k \binom{n}{k} k^p$ is the sum over maps $f: [p] \to n$ of the sum over subsets $S$ of $[n]$, containing the image of $f$, of size at most $k$, of $(1)^{S}$. We may assume the image of $f$ has size $\leq d < n $ and thus that there is some element $e$ not in the image of $f$. We can cancel each subset with $e\notin S$ with the $S \cup \{e\}$, as these have opposite signs. The only subsets that fail to cancel are those that have size exactly $d$ and do not contain $e$, of which there are $\binom{n   \operatorname{Im}(f)  1}{ d  \operatorname{Im}(f) } $.
With $S_2(p,j)$ again the Stirling numbers of the second kind, the number of maps from $[p]$ to $[n]$ with image of size $j$ is $ \frac{n!}{ (nj)!} S_2(p,j) $, so the sum is
$$ (1)^d \sum_{j=0}^d S_2(p,j) \frac{n!}{(nj)!} \binom{ nj1}{dj} $$
$$= (1)^d \frac{n!}{ (n1d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(nj)} \frac{1}{(dj)!} $$
(If $d=n$ then all subsets cancel and so only the terms with $ \operatorname{Im} f =n$ remain, so we just obtain the count of surjections from $[p]$ to $[n]$, as in Carlo Beenakker's answer.)
Alternately, a formulabased proof:
we have $$ k^p = \sum_{j=0}^k S_2( p,j) \frac{k!}{ (kj)!} $$ ( a standard identity.) so
$$\sum_{k=0}^d (1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (1)^k S_2( p,j) \frac{k!}{(kj)!} {n \choose k} $$ and $$\frac{k!}{(kj)!}{n\choose k} = \frac{k! n!}{ (kj)! k! (nk)! } = \frac{n!}{ (kj)! (nk)!} = \frac{n!}{(nj)!} \binom{nj}{kj} $$ so $$ \sum_{k=0}^d (1)^k k^p {n \choose k} = \sum_{j=0}^d \sum_{k=j}^d (1)^k S_2( p,j) \frac{n!}{(nj)!} \binom{nj}{kj}$$ $$ = \sum_{j=0}^d (1)^d S_2( p,j) \frac{n!}{(nj)!} \binom{nj1}{dj} = (1)^d \frac{n!}{ (n1d)!} \sum_{j=0}^d S_2(p,j) \frac{1}{(nj)} \frac{1}{(dj)!} $$

$\begingroup$ Thank you! It is at times like these that I wish I could accept two answers. $\endgroup$ Aug 31 '20 at 11:12
Up to the factor $(1)^n$, the uncut sum is $$s_{p,n}:=\sum_{k=0}^n(1)^{nk}\, k^p\,\binom nk.$$ 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, $$s_{p,n}=n!\binom pn ES_n^{pn} \tag{1}$$ 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, $$0<s_{p,n}\le n!\binom pn n^{pn}. \tag{2}$$
If $pn\ge1$, then, in view of Jensen's inequality, the lower bound $0$ on $s_{p,n}$ in (2) can be greatly improved, to $$b_{p,n}:=n!\binom pn \Big(\frac n2\Big)^{pn}.$$
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 $pn\to a$ for some real $a>0$, then $$s_{p,n}\sim b_{p,n}.$$
You can find plenty of documentation on Gould's site. Maybe it could be useful. The link is https://math.wvu.edu/~hgould/ Interesting files are Vol.1.PDF to Vol. 8.PDF.