I am trying to find a closed form expression (or a more easily computable expression) for the sum
$\sum_{k=1}^m \left({m\brace{k-1}} \cdot (n)_k\right)$,
where ${m\brace{k-1}}$ denotes a Stirling number of the second kind (number of ways of partitioning $n$ objects into $k-1$ nonempty sets) and $(n)_k := n(n-1)\cdots (n-k+1)$. Using the recurrence formula ${m+1\brace{k}} = k{m\brace{k}} + {m\brace{k-1}}$, we could try to evaluate
$\sum_{k=1}^m \left(k \cdot {m\brace{k}} \cdot (n)_k\right)$
instead. I was wondering if anyone has any ideas on how to simplify these sums?
Thank you!