To finish a proof i have been working on i must prove the following: $$ \sum_{k=0}^{n-1}\frac{n(-1)^{n-k+1}}{n-k}{n-1 \choose k} = 1 $$
I have checked that it does work empirically, but of course that is not good enough. Any suggestions on strategies for proving this summation would be just as appreciated, Thank you.
$$\require{cancel}\begin{align} \sum_{k=0}^{n-1}\frac{n}{n-k}{n-1\choose k}(-1)^{n-k+1}&= -\sum_{k=0}^{n-1}\frac{n}{\cancel{n-k}}\left[\frac{(n-1)(n-2)\cdots \cancel{(n-k)}}{1\cdot 2\cdots k}\right](-1)^{n-k}\\ &=-\sum_{k=0}^{n-1}{n\choose k}(-1)^{n-k}\\ &=-\sum_{k=0}^{n-1}{n\choose n-k}(-1)^{n-k}\qquad\text{using symmetry}\\ &=-\sum_{r=1}^{n}{n\choose r}(-1)^r\qquad\qquad \text{where $r=n-k$}\\ &=-\bigg\lbrace\left[\sum_{r=0}^{n}{n\choose r}(-1)^r\right]-{n\choose 0}\bigg\rbrace\\ &=-\left[(1-1)^n-1\right]\\ &=1\qquad \blacksquare\end{align}$$