Is there a nice combinatorial proof of the following identity? (That is, by showing that both sides count the same thing.) $$\sum_{k=0}^n k \cdot k! = (n+1)! -1 $$ I was searching Wikipedia for nice identities to assign to my students for a homework on combinatorial proof, and thought this one looked innocent enough, but then realized I couldn't solve it myself.
Perhaps we should take the set of permutations of $n+1$ letters and partition it in some clever way (maybe the $-1$ suggests that the identity should be set aside), but I can't see what that might be.
Of course it is very easy to prove by induction; that's not what I'm looking for.