$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where

$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$
Here $x_j=j/(n+1)$.

So, working through the algebra:

$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots
  =\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}
 $$

Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$
which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.
My question: How does one see that it equals 1?

$$
\frac{1}{1-e^{-a}} = \frac{1}{1-e^{-a}}\cdot\frac{e^a}{e^a} = \frac{e^a}{e^a-1} = \frac{-e^a}{1-e^a}.
$$
Hence
$$
\frac{1}{1-e^a} + \frac{1}{1-e^{-a}}  = 1.
$$
