Combinatorial argument for Stirling number identity

Question

define $S(n,k)$ to be the number of partitions of $\{1,2,3,...,n\}$ into $k$ nonempty subsets. Prove that for any $n \geq k$: $$kS(n,k) = \displaystyle \sum_{m=0}^{n-1}{n \choose m}S(m,k-1)$$ What I've come up with so far is an argument about putting books into book boxes so that the order of the books in the book boxes does not matter (the books are distinct and the book boxes are not). Inside the sum, we pick $m$ books from the $n$ in ${n \choose m}$ ways and we put those $m$ books into the first $k-1$ boxes in $S(m,k-1)$ ways, then place the remaining $n-m$ books into the $k^{th}$ box in exactly $1$ way. However, some problems I've noticed are that if $m < k-1$ then some of the boxes would be empty. Does anyone have a nice argument for this identity, any help would be greatly appreciated.

Answer 1

Hint: $kS(n,k)$ counts the number of ways to put $n$ objects into $k$ identical boxes, and then single out one of the boxes to be special. What if you first choose the items to go in the special box, and then partition the other objects into the remaining boxes?