Encouraged and emboldened (embiggened?) by the ingenuity displayed by some of our readers, I’d like to see what sort of response we get to this Problem of the Week:

Establish the following identity: $\displaystyle \sum_{j=1}^n \binom{n}{j} j^{j-1} (n-j)^{(n-j)} = n^n$ for all natural numbers $n > 0$.

(Here we make the convention $0^0 = 1$.) I find this problem tantalizing because it looks as if there should be some sort of conceptual proof — can you find one?

Please send your solutions to topological[dot]musings[At]gmail[dot]com by Wednesday, June 11, 11:59pm (UTC); do not submit solutions in Comments. Everyone with a correct solution gets entered in our Hall of Fame! We look forward to your response.