Stirling-like sum equal to zero when $k>n$

Question

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)

Answer 1

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.

Answer 2

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.