A question about summation formula involving binomial coefficient In Table of Integrals, Series, and Products. Seventh Edition. I.S. Gradshteyn and I.M. Ryzhik, there is 
0.154.3
$$ 
\sum_{k=0}^N (-1)^k {N \choose k}  k^{n-1} =0, N \geq n \geq 1; 0^0 ≡ 1
$$
0.154.4
$$
\sum_{k=0}^n (-1)^k {n \choose k}  k^{n} =(-1)^n n!,  n \geq 0; 0^0 ≡ 1
$$
I would like to know, there is any generalization for 0.154.4 for a general, higher power of $k$?
$$
\sum_{k=0}^n (-1)^k {n \choose k}  k^{n+m} =?,  n \geq 0; m\geq0; 0^0 ≡ 1
$$
 A: $$
\sum_{k=0}^n (-1)^k {n \choose k}  k^{n+m} = (-1)^n\cdot n!\cdot S(n+m,n),
$$
where $S(\cdot,\cdot)$ is Stirling number of the second kind. This is essentially formula (10) at http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html
In particular, for $m<0$, we have $S(n+m,n)=0$; and for $m=0$, we have $S(n+m,n)=S(n,n)=1$, giving the formulae 0.154.3-4 from Gradshteyn and Ryzhik.
A: Here is one way of looking at the answer to this problem. 
For any function $f: \mathbb{N} \to \mathbb{R}$, one can form the exponential generating function or formal power series $E_f(x) = \sum_{j \geq 0} \frac{f(j) x^j}{j!}$. The power series product of such egf's corresponds to the convolution product defined by 
$$(f \ast g)(n) = \sum_{j+k = n} \frac{n!}{j!k!} f(j) g(k).$$ 
For example, if $g(k) = (-1)^k$, then $E_g(x) = \exp(-x)$, and the convolution product 
$$h(n) := (f \ast g)(n) = \sum_j (-1)^{n-j} {n \choose j} f(j)$$ 
corresponds to the equation $E_h = E_f \cdot \exp(-x)$. Rewriting this as $E_h \exp(x) = E_f$ (and with an option to swap out other dummy variables for $j$), we arrive at a Möbius inversion formula 
$$h(n) = \sum_k (-1)^{n-k} {n \choose k} f(k) \qquad \Leftrightarrow \qquad f(k) = \sum_j {k \choose j} h(j).$$
In the above problem, we are thus, modulo a sign $(-1)^n$, trying to solve for the (unique) $h(j)$ satisfying 
$$k^{m+n} = \sum_j {k \choose j} h(j),$$ 
but notice that we can interpret $k^{m+n}$ combinatorially as the number of functions from an $(m+n)$-element set to a $k$-element set. Each such $f$ can be uniquely factored as a surjection followed by a subset inclusion; here ${k \choose j}$ counts the number of $j$-element subsets, and correspondingly $h(j)$ must be counting the number of surjections from an $(m+n)$-element set to a $j$-element set $[j] := \{1, 2, \ldots, j\}$. Such a surjection can be regarded as a partition of $[m+n]$ into $j$ nonempty classes together with a labeling of the classes by the elements $1, \ldots, j$, and there are $j!$ such labelings. The second Stirling number $S(m+n, j)$ is in fact combinatorially defined as the number of partitions on $[m+n]$ with $j$ classes, and so we have shown 
$$h(j) = S(m+n, j) \cdot j!$$ 
or $h(n) = S(m+n, n) \cdot n!$, as indicated in the answer above. 
A: Operationally, using the generalized Dobinski formula in the MO-Q on the o.g.f. for the Bell polynomials $\phi_m(x) = \sum_{k=0}^m S(m,k) x^k$,
$$ \phi_{n+m}(x) = (\phi.(x))^{n+m} = e^{-x} \sum_{j =0}^{\infty} j^{n+m} \frac{x^j}{j!} = \sum_{j=0}^{\infty} (-1)^j [\sum_{k=0}^j (-1)^k \binom{j}{k} k^{n+m}] \frac{x^j}{j!} \; . $$
Equivalently,
$$ \phi_{p}(\widehat{xD_x})= \sum_{k=0}^p S(p,k)x^kD_x^k = (xD_x)^p=\sum_{j=0}^\infty j^p \frac{x^jD^j_{x=0}}{j!}=\sum_{j=0}^\infty (-1)^j [\sum_{k=0}^j(-1)^k \binom{j}{k}k^p] \frac{x^jD_x^j}{j!} \; ,$$
where $D_x = \frac{d}{dx}$.
