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.
 A: Applying the same technique as per this answer
$$Q(x)=\prod\limits_{k=1}^n(x-x_k) \text{ and }
Q'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$
then 
$$\frac{Q'(x)}{Q(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$
or 
$$-\frac{Q'(1)}{Q(1)}=\sum\limits_{i=1}^{n}\frac{1}{x_i-1}$$
But $Q(1)=n+1$ and $Q'(x)=1+2x+3x^3+...+nx^{n-1}\Rightarrow Q'(1)=\frac{n(n+1)}{2}$ and
$$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$
A: $1$ and $x_1,\ldots,x_n$ are the roots of $z^{n+1}-1=0$. They are
the $(n+1)$-th roots of unity.
Then $1/(z^{n+1}-1)$
has a partial fraction expansion of the form
$$\frac1{z^{n+1}-1}=\frac a{z-1}+\sum_{k=1}^n\frac{b_k}{x_kz-1}.$$
Multiplying by $z-1$ and setting $z=1$ gives $a=1/(n+1)$.
Multiplying by $x_kz-1$ and setting $z=1/x_k$ gives $b_k=1/(n+1)$.
Therefore
$$\sum_{k=1}^n\frac1{x_kz-1}=\frac{n+1}{z^{n+1}-1}-\frac1{z-1}.$$
Letting $z\to1$ gives
$$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1}
=\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$
ADDED IN EDIT.
Here's a trick proof. Let $S$ denote the sum in question. As the
reciprocals of the $x_k$ are the $x_k$ again (in a different order) then
$$S=\sum_{k=1}^n\frac1{x_k^{-1}-1}=\sum_{k=1}^n\frac{x_k}{1-x_k}
$$
and
$$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right)
=\sum_{k=1}^n(-1)=-n.$$
A: We have $$x^{n+1}-1=0\ \ \ \ (1)$$ with $x\ne1$
Set $\dfrac1{x-1}=y\iff x=\dfrac{y+1}y$
Replace the value of $x$ in terms of $y$ in $(1)$ to form an $n$ degree equation in $y$ $$\left(\dfrac{y+1}y\right)^{n+1}=1$$
$$\implies\binom{n+1}1 y^n+\binom{n+1}2 y^{n-1}+\cdots+1=0$$
Now apply Vieta's formula to find $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$
A: Simply note that $1,x_1,\ldots, x_n$ are roots of $$
x^{n+1} -1 = 0.$$ Thus we have $0,x_1-1,\ldots, x_n-1$ are roots of
$$
(x+1)^{n+1} -1 = x\sum_{j=0}^{n} \binom{n+1}{j+1}x^j = 0.
$$
Hence,  $z_i = x_i -1$, $1\leq i \leq n$ are roots of
$$
\sum_{j=0}^{n} \binom{n+1}{j+1}z^j = 0,
$$ and by Vieta's formula, 
$$
\sum_{i=1}^n \frac{1}{z_i} = \frac{\sum_{i=1}^n z_1z_2\cdots z_{i-1}\widehat{z_i}z_{i+1}\cdots z_n}{z_1z_2\cdots z_{n-1}z_n} = -\frac{\binom{n+1}{2}}{\binom{n+1}{1}} = -\frac{n}{2}, 
$$ where $\widehat{z_i}$ means the variable is omitted in the calculation.
A: First, the denominator is, I think, $(-1)^nQ(1)$. 
Here is a hint for the numerator: how can you write $Q’(x)$?
A: 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} $.
A: We solve  it using  complex residues  by way  of enrichment.  With the
function
$$f(z) = \frac{1}{z-1} \frac{(n+1)/z}{z^{n+1}-1}$$
we have for $\zeta_k = \exp(2\pi i k/(n+1))$ with $1\le k\le n$ that
$$\mathrm{Res}_{z=\zeta_k} f(z)
= \frac{1}{z-1}
\left.\frac{(n+1)/z}{(n+1)z^n}\right|_{z=\zeta_k}
= \frac{1}{\zeta_k-1}.$$
Residues sum to zero so we have
$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1}
+ \mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=1} f(z)
+ \mathrm{Res}_{z=\infty} f(z) = 0.$$
The residue at infinity is zero by inspection. The residue at $z=0$ is
$n+1$, also by inspection. We have for the residue at $z=1$
$$\mathrm{Res}_{z=1} f(z) =
\mathrm{Res}_{z=1} \frac{1}{(z-1)^2}
\frac{(n+1)/z}{1+z+\cdots +z^{n}}
\\ = \left.\left(
\frac{(n+1)/z}{1+z+\cdots +z^{n}}
\right)'\right|_{z=1}
\\ = \left.\left(
- \frac{(n+1)/z^2}{1+z+\cdots +z^{n}}
- \frac{(n+1)/z \times (1+2z+\cdots+nz^{n-1})}
{(1+z+\cdots +z^{n})^2}
\right)\right|_{z=1}
\\ = -1 - \frac{(n+1) \times \frac{1}{2} n (n+1)}{(n+1)^2}.$$
Now with
$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1} =
- \mathrm{Res}_{z=0} f(z)
- \mathrm{Res}_{z=1} f(z)$$
we obtain
$$-(n+1) + 1 + \frac{1}{2} n.$$
and therefore our answer is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=1}^{n} \frac{1}{\zeta_{k}-1}
= - \frac{n}{2}.}$$
