Problem based on sum of reciprocal of $n^{th}$ roots of unity 
Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$

$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$
Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$
Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$
So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-......\right\}-y^n=0$
So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+...+(-1)^n=0$  has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,.......,y_{n-1}$
So $$y_{1}+y_{2}+y_{3}+.....+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = \frac{1}{1-x_{i}}\;\forall i\in \left\{1,2,3,4,5,.....,n-1\right\}$
My Question is can we solve it any less complex  way, If yes Then plz explain here, Thanks

Let $f(x) = \frac{(x^n-1)}{(x-1)} = x^{n-1}+x^{n-2}+...+x+1$
Then $\frac{f'(x)}{f(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2}+ ... + \frac{1}{x-x_{n-1}}$
So what you want is just $\frac{f'(1)}{f(1)}$, which is easy to compute as $\frac{n-1}{2}$.
