Closed form for series $\sum_{m=1}^{N}m^n\binom{N}{m}$ 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
$$
I_N(1)=N2^{N-1}; I_N(2)=N(N+1)2^{N-2}; I_N(3)=N^2(N+3)2^{N-3}
$$
 A: Suppose we are trying to evaluate
$$I_N(n) = \sum_{m=1}^N {N\choose m} m^n.$$
Observe that
$$m^n = \sum_{q=0}^m {n\brace q} \frac{m!}{(m-q)!}.$$
Note also that ${n\brace 0} = 0$ so that we get for the sum
$$\sum_{m=0}^N {N\choose m} 
\sum_{q=0}^m {n\brace q} \frac{m!}{(m-q)!}.$$
Re-write this as
$$\sum_{q=0}^N {n\brace q}
\sum_{m=q}^N {N\choose m} \frac{m!}{(m-q)!}$$
or
$$\sum_{q=0}^N {n\brace q} \times q! \times
\sum_{m=q}^N {N\choose m} {m\choose q}.$$
We now see by inspection (i.e.  considering subsets of size $q$ of $N$
elements) that the inner sum can be simplified to give
$$\sum_{q=0}^N {n\brace q} \times q! \times
{N\choose q} 2^{N-q}.$$
Now it remains  to show how to compute the  Stirling numbers for fixed
$q.$ Recall the marked species of set partitions
$$\mathfrak{P}(\mathcal{U}(\mathfrak{P}_{\ge 1}(\mathcal{Z})))$$
which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1))$$
and hence
$${n\brace q} = n! [z^n] \frac{(\exp(z)-1)^q}{q!}.$$
Suppose we wanted to compute $I_N(3).$
We get
$${n\brace 1} = n! [z^n] \frac{\exp(z)-1}{1!}
= n! \frac{1}{n!} = 1$$
and
$${n\brace 2} = n! [z^n] \frac{(\exp(z)-1)^2}{2!}
= \frac{n!}{2!} \times 
\left(\frac{2^n}{n!}-2\frac{1}{n!}\right)
= 2^{n-1} - 1.$$
and finally
$${n\brace 3} = n! [z^n] \frac{(\exp(z)-1)^3}{3!}
= \frac{n!}{3!} \times 
\left(\frac{3^n}{n!} - 3\frac{2^n}{n!}
+ 3\frac{1}{n!}\right)
\\ = \frac{1}{6} 3^n
- \frac{1}{2} 2^n + \frac{1}{2}.$$
This gives for $I_N(3)$ the expression
$${N\choose 1} 2^{N-1}
+ (2^2-1) \times 2 \times {N\choose 2} 2^{N-2}
+ \left(\frac{1}{6} 3^3 - \frac{1}{2} \times 2^3 + \frac{1}{2}\right)
\times 6 \times {N\choose 3} 2^{N-3}.$$
This is
$$2^{N-3} \times
\left(4N + 6N(N-1)+ N(N-1)(N-2)\right).$$
which simplifies to
$$I_N(3) = N^2\times (N+3)\times 2^{N-3}.$$
A: (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,
$$I_N(n)=\sum_{m=1}^N{n\brace m}N^{\underline m}2^{N-m}=\sum_{m=1}^N{n\brace m}m!\binom{N}m2^{N-m}\;.$$
