Evaluate $\sum_{k = 0}^{n} {n\choose k} k^m$ [duplicate]
Solution 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:
$$ D^s (1+x)^n = \frac{\Gamma(n+1)}{\Gamma(n-s+1)}(1+x)^{n-s} .$$