Proving $\sum_{j=0}^{n}\binom{n}{j}j^{n-2}(-1)^{n-j+1} = 0$

Question

Prove $$\sum_{j=0}^{n}\binom{n}{j}j^{n-2}(-1)^{n-j+1} = 0$$

My wrong proof try is this:

$\sum_{j=0}^{n}\binom{n}{j}j^{n-2}(-1)^{n-j+1} = \sum_{j=0}^{n}\binom{n}{j}1^jj^{n-2}(-1)^{n-j}(-1) =-\sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2}$

We know:

$(\forall x,y \in R)(\forall n \in N) (x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$

Therefore:

$\sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j} = (1-1)^n = 0$

And

$\sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2} \leq \sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}n^{n-2} \\ = n^{n-2}\sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j} \\ = 0$

And

$ \sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2} = \sum_{j=1}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2} \\ \geq \sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}1^{n-2} \\ = 1\sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j} \\ = 0$

Therefore

$0 \leq \sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2} \leq 0 \Rightarrow \\ \sum_{j=0}^{n}\binom{n}{j}1^j(-1)^{n-j}j^{n-2} = 0$

And

$\sum_{j=0}^{n}\binom{n}{j}j^{n-2}(-1)^{n-j+1} = 0$

I later realised that the comparison cant be made due to the $(-1)^n$

I have also tried by testing for n even and odd numbers. I can prove it for odd numbers but not for even.

Do you have any other ideas?

Answer 1

I fully expect this will be marked a duplicate since it has appeared so many times. We have

$$\sum_{j=0}^n {n\choose j} j^{n-2} (-1)^{n-j} = (n-2)! [z^{n-2}] \sum_{j=0}^n {n\choose j} \exp(jz) (-1)^{n-j} \\ = (n-2)! [z^{n-2}] (\exp(z)-1)^n.$$

Now since $\exp(z)-1 = z + \cdots$ we obtain $(\exp(z)-1)^n = z^n + \cdots$ so that $(n-2)! [z^{n-2}] (\exp(z)-1)^n = (n-2)! [z^{n-2}] (z^n+\cdots) = 0.$ This is for $n\ge 2.$