Stirling-like sum equal to zero when $k>n$ I need to prove that 
$$\sum_{r=0}^k\binom{k}{r}(-1)^r r^n=0$$
when $n<k$.
I know that the formula above can be easily transformed into the Stirling number of the Second kind formula, which is derived from combinatorics (number of ways to split $n$ objects into $k$ groups) meaning it must be $0$ when $n<k$. I'd like to see if one can prove this without using combinatorics.
I haven't tried much since I have no idea where to start, any suggestion would be welcome.
(The problem arised while doing some calculations regarding Bernoulli numbers)
 A: Related technique: (I). Here is an approach. Recalling the identities

$$ \sum_{r=0}^k\binom{k}{r}x^r = (1+x)^k $$
$$ (xD)^n = \sum_{m=0}^{n} {n\brace m} x^mD^m  $$

where $D=\frac{d}{dx}$ and $ {n\brace m} $ is the Stirling numbers of the second kind. Applying the operator $(xD)^n$ to both sides of the first identity gives
$$ (xD)^n\sum_{r=0}^k\binom{k}{r}  x^r =  \sum_{r=0}^k\binom{k}{r} r^n x^r = (xD)^n (1+x)^k .$$
I think you can finish the problem. Remember that you need to substitute $x=-1$ at the end. Note that if you substitute $x=-1$ in the first identity you will get zero.
A: Observe that
$$r^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp(rz) \; dz.$$
This gives for the sum that
$$\sum_{r=0}^k {k\choose r} (-1)^r r^n
= \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\sum_{r=0}^k {k\choose r} (-1)^r \exp(rz) \; dz$$
This is
$$\frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} 
\left(1-\exp(z)\right)^k \; dz.$$
Note however that 
$$1-\exp(z) = -\frac{z^1}{1!} -\frac{z^2}{2!} -\frac{z^3}{3!}-\cdots$$
Therefore $(1-\exp(z))^k$ starts at $[z^k]$ and hence when $k\gt n$ the integrand is an entire function and the integral vanishes.
