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.

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
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=1}^{n} \frac{1}{\zeta_{k}-1}
= - \frac{n}{2}.}$$
