Proving $\sum\limits_{r=1}^n \cot \frac{r\pi}{n+1}=0$ using complex numbers Let $x_1,x_2,...,x_n$ be the roots of the equation $x^n+x^{n-1}+...+x+1=0$.

The question is to compute the expression $$\frac{1}{x_1-1} + \frac{1}{x_2-1}+...+\frac{1}{x_n-1}$$
  Hence to prove that $$\sum_{r=1}^n \cot \frac{r\pi}{n+1}=0$$ 

I tried rewriting the expression as $$\sum_{i=1}^n \frac{\bar{x_i}-1}{|x_i-1|^2}$$
I then used the fact that $$x^{n+1}-1=(x-1)(x^n+x^{n-1}+...+x+1=0$$ so $x_i$ are the complex nth roots of unity.Using cosine formula I found that $$|x_i-1|^2=2-2\cos(\frac{2i\pi}{n+1})=4(\sin \frac{\pi}{n+1})^2$$
After substituting this I couldn't simplify the resulting expression.Any ideas?Thanks.

Clearly, $x_k\ne1,1\le k\le n$
Let $y_k=\dfrac1{x_k-1}\ne0,1\le k\le n$
$\implies x_k=\dfrac{1+y_k}{y_k}$
Now $\displaystyle0=\sum_{r=0}^nx_k^r=\dfrac{x^{n+1}_k-1}{x_k-1}$
As $x_k-1$ is non-zero finite, $$x^{n+1}_k-1=0$$
$$\implies\left(\dfrac{1+y_k}{y_k}\right)^{n+1}=1$$
$$\binom{n+1}1y_k^n+\binom{n+1}2y_k^{n-1}+\cdots+1=0$$
$$\sum_{r=1}^ny_k=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$
Now as $x_k=e^{2k\pi i/(n+1)},1\le k\le n$
$$y_k=\dfrac1{e^{2k\pi i/(n+1)}-1}=\dfrac{e^{-k\pi i/(n+1)}}{e^{k\pi i/(n+1)}-e^{-k\pi i/(n+1)}}$$
$$=\dfrac{\cos\dfrac{k\pi}{n+1}-i\sin\dfrac{k\pi}{n+1}}{2i\sin\dfrac{k\pi}{n+1}}=?$$
Used: How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?
$\implies e^{-it}=\cos t-i\sin t$
