We denote Stirling numbers of the first kind as $s(n,m)$ counting the ways we can put $n$ people in $m$ cycles.
What is a combinatorial/algebraic proof to this identity?
$$s(n+1,m+1)=\sum_k{s(n,k). ^kC_m}$$
Which $^kC_m$ is $ k$ choose $m$.
I tried to say like this: we put one person away so we have $n$ people left. Then we put them around $k\geq m$ cycles. Then we choose $m$ of these cycles and hold them as they are. Then merge the people on other cycles and the person we picked at the beginning. But I can't explain this merging. Maybe my approach isn't true from the beginning. Any help is appreciated!
