Evaluate $\sum_{k = 0}^{n} {n\choose k} k^m$

Question

So, I wonder what is the evaluation of $$\sum_{k = 0}^{n} {n\choose k} k^m\text{,}\qquad (*)$$ where $m,n\in \mathbb{N}$.

One of my tries: knowing that $$k^m = \sum_{j = 0}^{m}\text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{,}$$ where $\text{S}(m,k)$ are Stirling numbers of the second kind, and $$\sum_{k = 0}^n {n\choose k} k(k-1)\cdots(k-j+1) =2^{n-j}\cdot n(n-1)\cdots(n-j+1)\text{,}$$ I have rewritten the upper sum into $$\sum_{k = 0}^{n}\sum_{j = 0}^m {n\choose k} \text{S}(m,j)\cdot k(k-1)\cdots(k-j+1)\text{.}$$ Changing the order of summation, we get $$\sum_{j = 0}^{m}\text{S}(m,j) \cdot 2^{n-j}\cdot n(n-1)\cdots (n-j+1)\text{,}$$ and here it stops.

Answer 1

related techniques: (I), (II). Here is how you advance. Recalling the identities

$$ \sum_{k = 0}^{n} {n\choose k} x^k = (1+x)^n, $$

and

$$ (xD)^m = \sum_{s=0}^{m} {m\brace s} x^s D^s, $$

where $D=\frac{d}{dx}$ and ${m\brace s}$ are Stirling numbers of the second kind. Now, apply the above operator to both sides of the first equation as

$$ (xD)^m \sum_{k = 0}^{n} {n\choose k} x^k = \sum_{s=0}^{m} {m\brace s} x^s D^s (1+x)^n. $$

I leave it for you to finish the problem.

Note:

$$ D^s (1+x)^n = \frac{\Gamma(n+1)}{\Gamma(n-s+1)}(1+x)^{n-s} .$$