While computing conditional expectations of certain functionals of a Poisson white noise field (details are long and probably irrelevant), I've stumbled upon the need to use the following identity involving Stirling numbers of the second kind:
$$
\ell{k\brace \ell} = \sum\_{j=\ell}^k {k\choose j-1} (-1)^{k-j} {j\brace \ell}.
$$
I used Manuel Kauers' [*Stirling*](http://www.kauers.de/software.html) package in order to produce a recurrence relation from which the identity can be easily proved. I still wonder, however, whether this is actually well-known, or there is some short proof...



Proof of identity involving Stirling numbers of the second kind
You can give a short proof by interpreting the identity as an instance of inclusion-exclusion. The left hand side counts the number of ways of partitioning $S=\{1,2,\dots,k\}$ into $\ell$ parts and then picking one of the parts as the *designated* one.


Let $A\_i$ denote the number of partitions of $S$ into $\ell$ parts where the designated part contains $i$. You can check that the left hand side is counting $|A\_1\cup A\_2\cup \cdots \cup A\_k|$. For the right hand side notice that 
$$|A\_{i\_1}\cap A\_{i\_2}\cap\cdots\cap A\_{i\_r}|={k-r+1\brace \ell}$$
So by inclusion-exclusion we get
$$|A\_1\cup A\_2\cup \cdots \cup A\_k|=\sum\_{r=1}^{k-l+1}(-1)^{r-1}\binom{k}{r}{k-r+1\brace \ell}$$ 
and reindexing by $j=k-r+1$ gives your identity.


