<s>[INST] If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$ 
Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of
  $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$

Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)
My attempt:
Developing expression $P(1)$, replacing the 1 by $x$, follows
$$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$ 
As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that 
$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$
and 
$$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$
Therefore the denominator of $P(1)$ is 
$$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.
Another fact that is probably useful is that 
$$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$
with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is
$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$
This is as far as I could go...  
Hints and full answers are welcomed.
 [/INST] We notice that the equation:
$$x^{n} + x^{n-1} + ... +x +1 = 0 $$
is a geometric sequence and can be rewritten as:
$$x^{n} + x^{n-1} + ... +x +1 = \sum_{i=0}^{n}x^n = \frac{1-x^{n+1}}{1-x} = 0 $$
From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from:
$$ 1-x^{n+1} = 0 \rightarrow x^{n+1} = e^{2\pi k} $$
$$ x_k = e^{2\pi k/(n+1)} $$
Where $k \ne 0$ and goes from $1$ to $n$. Because $k\ne 0$ then $x\ne1$ and there is no problem with the denominator anymore. The sum becomes:
$$ \sum_{k=1}^{n} \frac{1}{1-x_k} = \sum_{k=1}^{n} \frac{1}{1-e^{2\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{e^{-\pi k/(n+1)}}{e^{-\pi k/(n+1)}-e^{\pi k/(n+1)}} = \sum_{k=1}^{n} \frac{\cos(\pi k/(n+1))-i\sin(\pi k/(n+1))}{-2i\sin(\pi k/(n+1))} = $$
$$ \sum_{k=1}^{n} \frac{1}{2} + i\frac{\cot(\pi k/(n+1))}{2} = \frac{n}{2} + \frac{i}{2}\sum_{k=1}^{n}\cot(\pi k/(n+1)) = \frac{n}{2} $$
From Wolfram alpha $ \sum_{k=1}^{n}\cot(\pi k/(n+1)) = 0 $ but you can prove it easily. An example:
$$ \cot(\frac{n\pi}{n+1}) = \cot(\frac{(n+1)\pi-\pi}{n+1}) = \cot(\pi- \frac{\pi}{n+1}) = -\cot(\frac{\pi}{n+1})  $$
which will cancel out with the first term and so on with other terms.
NOTE: I conducted the calculations for the sum $ \sum_{k=1}^{n} \frac{1}{1-x_k} $.
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