$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$? 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?

To see that the given expression is equal to $1$, set
$\omega = e^{\frac{ik\pi}{n + 1}}, \tag{1}$
then the expression becomes 
$\dfrac{1}{1 - \omega} + \dfrac{1}{1 - \bar{\omega}} =  \dfrac{1 - \bar{\omega} + 1 - \omega}{(1 - \omega)(1 - \omega)} =  \dfrac{2 - (\omega + \bar{\omega})}{(1 - \omega)(1  - \bar{\omega})}$
$= \dfrac{2 - (\omega + \bar{\omega})}{(1 + \omega \bar{\omega}) - (\omega + \bar{\omega})} = \dfrac{2 - (\omega + \bar{\omega})}{2 - (\omega + \bar{\omega})} = 1, \tag{2}$
since $\omega \bar{\omega} = 1$.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
