Prove that $n^n = \sum^{n}_{k=1}$ $k!$ $n\choose k $$kn^{n-k-1}$ using a combinatorial argument
My attempt:
For the LHS I considered $n$ distinct people to be placed into $n$ distinct rooms each of which has no restriction on the number of people in it. So they can be placed in $n^n$ ways as each person has n choices for a room.
For the RHS, I let k be the number of rooms that are being occupied and so we choose k rooms from the total n in $n \choose k $ ways. This is where I am getting stuck, as the values of the terms that are attached to $n \choose k $ in the expression are not coming to be the same as when I calculate it explicitly.
For eg. say $n = 3$ and $k = 2$ then we can place the people in $3 \choose 2 $6 ways. But from the term in the expression we get
$2!$$3 \choose 2 $ 2 = 4$3 \choose 2 $ ways.
I also tried manipulating the terms in the expression to make it work for eg.
$\sum^{n}_{k=1}$ $k!$ $n\choose n-k $$kn^{n-k-1}$ or
$\sum^{n}_{k=1}$ $(n-k)!$ $n\choose k $$(n-k)n^{k-1}$
But none of them work... So I think the initial scenario for the combinatorial argument is wrong and I cannot think of any other
