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.
$$\sum\limits_{k=1}^{n} (-1)^{k+1}\frac{n \choose k}{k}= 1+\frac 1 2 + \frac 1 3 + \dots \frac 1 n \tag{1}$$
On
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.
\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.