# Combinatorial Proof of $\sum_{k=1}^{n-1} k\cdot k!= n!-1$ [duplicate]

For $n ≥ 2$, $$\sum_{k=1}^{n-1} k \cdot k!= n!-1$$

On the left-hand side, we could be choosing ordered subteams from $n-1$ people (let's say for some reason one of these people cannot be qualified). the range of k would be the size of subteams and $k!$ would be ordering them. $k$ could also be ${k \choose 1}$, which means we could be choosing a team leader for each ordered subteam. On the right-hand side, we order all $n$ people and then take out one of these cases but what I can't figure out is how to equate these two sides. Could you give any hints that could help me progress here?

Note: I apologize that this was a duplicate. Thanks for everyone who helped!

• – StubbornAtom Oct 11 '16 at 7:15

Number the players from $1$ to $n$ and ask them to stand in increasing order. For each $k$:
• Rearrange the first $k$ players in any of $k!$ ways.
• Insert player $k+1$ in front of any of the first $k$ players (in $k$ ways).
Every permutation of $n$ can be realized in this way except for the identity. I'll leave it to you to establish the bijection.
• @Mike Person number $k+1$ gets inserted into the ordering "in front of" one of the first $k$, and thus will not be at his original position. – Andreas Blass Oct 11 '16 at 5:18