The Stirling numbers of the second kind ${n\brace k}$ are the number of distributions of $n$ distinct balls into $k$ non-empty, indistinguishible boxes.
These numbers have besides some other interesting representations via generating functions the `vertical' GF
\begin{align*}
\sum_{n=k}^\infty {n\brace k}\frac{t^n}{n!}=\frac{1}{k!}(e^t-1)^k\qquad\qquad k\geq 0
\end{align*}
A corresponding bivariate GF is
\begin{align*}
\sum_{k=0}^\infty\sum_{n=k}^\infty {n\brace k}\frac{t^n}{n!}u^k=\exp(u(e^t-1))\qquad\qquad
\end{align*}On the other hand there is the well known recurrence relation
\begin{align*}
{n\brace k}&=k{n-1\brace k}+{n-1\brace k-1}\qquad\qquad n,k\geq 1\tag{1}\\
{n\brace 0}&={0\brace k}=0\qquad\text{except }{0\brace 0}=1
\end{align*}
Note: A thorough treatment of Stirling numbers can be found e.g. in chapter 5 of Advanced Combinatorics by L. Comtet which also provides two proofs of the recurrence relation, a combinatorial one and an algebraic one based upon another generating function.
Since the information of the recurrence relation (1) is also encoded in the generating functions stated above, I've tried to derive it from them. Yet, I was not successful up to now.
So, any help to derive the recurrence relation from one of the generating functions above is highly appreciated.
Best Answer
$$\frac d{dt}\frac1{k!}(e^t-1)^k=\frac1{(k-1)!}(e^t-1)^{k-1}e^t=k(\frac1{k!}(e^t-1)^k)+\frac1{(k-1)!}(e^t-1)^{k-1}$$ Can you derive it from this?