Binomial identity $\sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{n-1 \choose k}{n-k} = \frac 2 n$ Prove that when $n$ is even:
$$\sum\limits_{k=1}^{n-1}(-1)^{k+1}\frac{n-1 \choose k}{n-k} = \frac 2 n$$
Note that when $n$ is odd, the identity above becomes $0$. It arose out of my attempt at proving equation (1) below (details here). Equation (1) was eventually proven via a different route, so I guess this makes for a long-winded proof, but I'm looking for a more direct way.
$$\sum\limits_{k=1}^{n} (-1)^{k+1}\frac{n \choose k}{k}= 1+\frac 1 2 + \frac 1 3 + \dots \frac 1 n \tag{1}$$
 A: 
We  obtain
  \begin{align*}
\color{blue}{\sum_{k=1}^{n-1}(-1)^{k+1}\frac{\binom{n-1}{k}}{n-k}}
&=\sum_{k=1}^{n-1}(-1)^{k+1}\frac{1}{n}\binom{n}{k}\tag{1}\\
&=-\frac{1}{n}\sum_{k=1}^{n-1}\binom{n}{k}(-1)^{k}\\
&=-\frac{1}{n}\left((1-1)^n-\binom{n}{0}-(-1)^n\binom{n}{n}\right)\tag{2}\\
&\,\,\color{blue}{=\frac{1}{n}\left(1+(-1)^n\right)}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the binomial identity $\binom{p}{q}=\frac{p!}{q!(p-q)!}=\frac{p}{p-q}\cdot\frac{(p-1)!}{q!(p-q-1)!}=\frac{p}{p-q}\binom{p-1}{q}$.

*In (2) we apply the binomial theorem.
A: \begin{align}
\sum_{k=0}^{n-1}\frac{(-1)^{k+1}x^{n-k}{n-1\choose k}}{n-k}&=\int_0^x\sum_{k=0}^{n-1}(-1)^{k+1}y^{n-k-1}{n-1\choose k}dy\\
&=-\int_0^x(y-1)^{n-1}dy\\
&=\frac{(-1)^n-(x-1)^n}{n}\ .
\end{align}
Substituting $\ x=1\ $ in this identity gives
\begin{align}
\sum_{k=0}^{n-1}\frac{(-1)^{k+1}{n-1\choose k}}{n-k}&=-\frac{1}{n}+\sum_{k=1}^{n-1}\frac{(-1)^{k+1}{n-1\choose k}}{n-k}\\
&=\frac{(-1)^n}{n}\ ,
\end{align}
from which the result follows.
