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 question about summation formula involving binomial coefficient
$$
\sum\_{k=0}^n (-1)^k {n \choose k} k^{n+m} = (-1)^n\cdot n!\cdot S(n+m,n),
$$
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.


