I'm currently struggling to find a way to derive an equivalent formula from the sum of two sums.
I'm looking at this problem $$ \sum_{k=0}^{n-1} (-1)^k \binom{n}{k} (n-k-1)! + (-1)^n + \sum_{k=0}^{n-2} (-1)^k \binom{n-1}{k} (n-k-2)! + (-1)^{n-1} = \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k}(n-k-1)! $$
I get as far as merging the right hand side into one sum $$ \sum_{k=0}^{n-2}\left(\binom{n}{k}(n-k-1)!+\binom{n-1}{k}(n-k-2)!\right)+(-1)^{n-1}n $$ $$ = \sum_{k=0}^{n-2}(-1)^k\left(\frac{n!}{k!(n-k)!}(n-k-1)! + \frac{(n-1)!}{k!(n-k-1)!}(n-k-2)!\right) + (-1)^{n-1} n $$ That's the point I am at right now. I would appreciate a tiny hint on how to proceed to derive the desired form.