Stirling numbers of the second kind - proof 2

Question

For a fixed integer $k$, how would I prove that

$\sum_{n\ge k} \left\{n \atop k \right\} \frac{x^n}{n!}=\frac{1}{k!}(e^x -1)^k $

where $\left\{n \atop k\right\} = k\left\{n-1 \atop k\right\}+\left\{n-1 \atop k-1\right\}$

Answer 1

In proving

$$\sum_{n\ge k} {n\brace k} \frac{z^n}{n!} = \frac{(\exp(z)-1)^k}{k!}$$

we use induction. We get for $k=1$

$$\sum_{n\ge 1} {n\brace 1} \frac{z^n}{n!} = \sum_{n\ge 1} \frac{z^n}{n!} = \frac{(\exp(z)-1)^1}{1!}$$

and the base case holds. For the induction step we have supposing it holds for $k$

$$\frac{\exp(z)-1}{k+1} \sum_{n\ge k} {n\brace k} \frac{z^n}{n!} = \frac{(\exp(z)-1)^{k+1}}{(k+1)!}.$$

Note however that for $m\ge k+1$ we get (power series starts at $z^{k+1}$)

$$[z^m] \frac{\exp(z)-1}{k+1} \sum_{n\ge k} {n\brace k} \frac{z^n}{n!} = \sum_{q=k}^{m-1} {q\brace k} \frac{1}{q!} [z^{m-q}] \frac{\exp(z)-1}{k+1} \\ = \frac{1}{m!} \times \frac{1}{k+1} \sum_{q=k}^{m-1} {m\choose m-q} {q\brace k}.$$

Next observe that

$$\sum_{q=k}^{m-1} {m\choose m-q} {q\brace k} = (k+1) {m\brace k+1}$$

because the LHS counts marked set partitions of $m$ into $k+1$ sets: we first choose $m-q$ elements of $[m]$ that go into the set bearing the marker, and partition the rest into $k$ sets. Clearly we obtain every partition of $m$ into $k+1$ sets (RHS) exactly $k+1$ times (available choices for the marker). This concludes the argument by induction because we have shown that

$$\frac{(\exp(z)-1)^{k+1}}{(k+1)!} = \sum_{m\ge k+1} {m\brace k+1} \frac{z^m}{m!}.$$