Setting. Let s(n,k) be the Stirling number of the first kind counting the number of permutations on a $n$-set with exactly $k$ cycles. For all non-negative integers $m,n$ there is the identity $$\sum_{k=0}^n s(n,k) m^k = m^{\overline{n}}$$ where $m^{\overline{n}} = m (m+1) \cdots (m+n-1)$ denotes the rising factorial. It is not hard to show the identity by induction using the recurrence formula $s(n,k) = s(n-1,k-1) + (n-1)s(n-1,k)$.
Question. Is there a nice combinatorial proof for this identity?
Motivation. For the Stirling numbers of the second kind, there is the somewhat similar identity $$\sum_{k=0}^n S(n,k) m^{\underline{k}} = m^n$$ involving the falling factorial. The common combinatorial proof is to identify both sides as the number of maps $f : N \to M$ (with $\#N = n$ and $\#M = m$), on the left hand side sorted wrt to the size $k = \#f(N)$ of the image.
