TITLE: Closed form for series $\sum_{m=1}^{N}m^n\binom{N}{m}$
QUESTION [5 upvotes]: How can we calculate the series
$$
I_N(n)=\sum_{m=1}^{N}m^n\binom{N}{m}?
$$
with $n,N$ are integers. 
The first three ones are
$$
I_N(1)=N2^{N-1}; I_N(2)=N(N+1)2^{N-2}; I_N(3)=N^2(N+3)2^{N-3}
$$

REPLY [3 votes]: (This isn’t really an answer, but it’s too long for a comment.) 
The first half of Marko’s answer can also be obtained by a simple combinatorial argument.
It’s not hard to see that $I_N(n)$ counts the ordered pairs $\langle f,A\rangle$ such that $f:[n]\to[N]$ and $\operatorname{ran}f\subseteq A$: for each $m\in[N]$, $m^n\binom{N}m$ is the number of pairs $\langle f,A\rangle$ such that $A\subseteq[N]$, $|A|=m$, and $\operatorname{ran}f\subseteq A$.
We can also count these pairs in the following way. For $m\in[N]$ there are $n\brace m$ partitions of $[n]$ into $m$ parts; these parts will be the fibres of a function $f:[n]\to[N]$. The function $f$ can be chosen in $N^{\underline m}=m!\binom{N}m$ ways. Then we can choose a subset $A$ of $[N]$ that contains $\operatorname{ran}f$ in $2^{N-m}$ ways. Thus,
$$I_N(n)=\sum_{m=1}^N{n\brace m}N^{\underline m}2^{N-m}=\sum_{m=1}^N{n\brace m}m!\binom{N}m2^{N-m}\;.$$