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Prove the identity $$(e^z-1)^m=m!\sum_{n}^{}{n \brace m}\frac{z^n}{n!}$$ $n\brace m$ stands for Stirling numbers of the second kind. I'm not really sure if $z$ is some special number or just an ordinary variable. And I have no clue on how to start.

Thanks!

Snape
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  • Do you know the generating function for the Stirling numbers of the second kind? If so, then this identity follows immediately upon expanding the outer exponential. If you can't use that, you'll need to start from some other characterization of the Stirling numbers; in that case, you should indicate whichever definition you're familiar with. – Semiclassical Aug 20 '14 at 16:18

1 Answers1

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Suppose you are allowed to use the OGF

$$F(z) = \sum_{n\ge m} {n\brace m} z^n = \prod_{r=1}^m \frac{z}{1-rz} = z^m \prod_{r=1}^m \frac{1}{1-rz}.$$

which seems like a reasonable assumption because it has a combinatorial proof.

Start by doing a partial fraction decomposition of the product term. We have with it being rational that $$\mathrm{Res}\left(\prod_{r=1}^m \frac{1}{1-rz}; z=\frac{1}{q}\right) = \prod_{r=1}^{q-1} \frac{1}{1-r/q} \times \left(-\frac{1}{q}\right) \times \prod_{r=q+1}^m \frac{1}{1-r/q} \\ = \left(-\frac{1}{q}\right) \times \prod_{r=1}^{q-1} \frac{q}{q-r} \times \prod_{r=q+1}^m \frac{q}{q-r} = - \frac{1}{q} q^{m-1} \frac{(-1)^{m-q}}{(q-1)! \times (m-q)!} \\ = - \frac{1}{m!} q^{m-1} (-1)^{m-q} {m\choose q}.$$

This gives for the OGF $F(z)$ that $$F(z) = -\frac{1}{m!} z^m \sum_{q=1}^m \frac{1}{z-1/q} q^{m-1} (-1)^{m-q} {m\choose q}.$$

In particular, $$[z^n] F(z) = {n\brace m} = -\frac{1}{m!} [z^n] z^m \sum_{q=1}^m \frac{1}{z-1/q} q^{m-1} (-1)^{m-q} {m\choose q} \\= \frac{1}{m!} [z^{n-m}] \sum_{q=1}^m \frac{q}{1-zq} q^{m-1} (-1)^{m-q} {m\choose q} \\ = \frac{1}{m!} \sum_{q=1}^m q^m q^{n-m} (-1)^{m-q} {m\choose q} = \frac{1}{m!} \sum_{q=1}^m q^n (-1)^{m-q} {m\choose q}.$$

To conclude introduce the EGF $$Q(z) = \sum_{n\ge 1} {n\brace m} \frac{z^n}{n!}$$ which is $$ \sum_{n\ge 1} \frac{z^n}{n!} \frac{1}{m!} \sum_{q=1}^m q^n (-1)^{m-q} {m\choose q}.$$

Interchanging the two summations this becomes $$ \frac{1}{m!} \sum_{q=1}^m (-1)^{m-q} {m\choose q} \sum_{n\ge 1} \frac{z^n q^n}{n!} = \frac{1}{m!} \sum_{q=1}^m (-1)^{m-q} {m\choose q} (\exp(qz)-1)$$

The term at $q=0$ is zero, so we get $$\frac{1}{m!} \sum_{q=0}^m (-1)^{m-q} {m\choose q} (\exp(qz)-1) = \frac{1}{m!} \sum_{q=0}^m (-1)^{m-q} {m\choose q} \exp(qz) \\ = \frac{1}{m!} (\exp(z)-1)^m.$$

The combinatorial proof of the OGF including some rather pedestrian work by this user can be found at this MSE link.

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