An identity containing $\exp\left(i\cdot \frac{2k\pi}{n} \right)$ I found the following equation without proof.
$$\frac{1}{(1+x)^n - 1} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{a(k,n)}{x + 1 – a(k,n)}$$
where $$a(k,n) = \exp\left(i \cdot \frac{2k\pi}{n} \right); \quad n = 1,2,\ldots; \quad k=0,1,2,\ldots,n-1.$$
Note that $(a(k,n))^n = 1$.
I tried to proof the above equation but failed. I’d appreciate it if you could help me out.
 A: With  the poles  of the  rational function  under consideration  being
simple and the numerator of lesser degree than the denominator we find
$$\frac{1}{(1+z)^n-1} =
\sum_{k=0}^{n-1}
\frac{1}{z-\rho_k}
\mathrm{Res}_{z=\rho_k}  \frac{1}{(1+z)^n-1}$$
where the $\rho_k$ are the roots of $(1+z)^n-1$ i.e.
$$\rho_k = \exp(2\pi i k/n) - 1.$$
Computing the residues we get
$$\sum_{k=0}^{n-1}
\frac{1}{z-\rho_k}
\frac{1}{n(1+\rho_k)^{n-1}}
= \sum_{k=0}^{n-1}
\frac{1}{z-\rho_k}
\frac{1+\rho_k}{n(1+\rho_k)^{n}}
\\ = \sum_{k=0}^{n-1}
\frac{1}{z-\rho_k}
\frac{1+\rho_k}{n}
= \frac{1}{n}
\sum_{k=0}^{n-1}
\frac{1+\rho_k}{z-\rho_k}.$$
Observing that $a(k,n) = 1 + \rho_k$ this becomes
$$\frac{1}{n}
\sum_{k=0}^{n-1}
\frac{a(k,n)}{z+1-a(k,n)}$$
as claimed.
A: We have the following polynomial with its roots:
$$P(x)= (x+1)^n-1=\prod\limits_{j=0}^{n-1}\left(x-x_j\right)$$
From which
$$P'(x)=n(x+1)^{n-1}=\sum\limits_{k=0}^{n-1}\prod\limits_{j=0,j\ne k}^{n-1}\left(x-x_j\right)$$
then 
$$\color{blue}{\frac{P'(x)}{P(x)}=\sum\limits_{k=0}^{n-1}\frac{1}{x-x_k}} \Rightarrow  \\
\frac{xP'(x)}{P(x)}=\sum\limits_{k=0}^{n-1}\frac{x}{x-x_k} \Rightarrow \\
\frac{xP'(x)}{P(x)}-n=\sum\limits_{k=0}^{n-1}\left(\frac{x}{x-x_k}-1\right) \Rightarrow \\
\frac{xP'(x)}{P(x)}-n=\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}$$
or
$$\frac{xn(x+1)^{n-1}}{(x+1)^n-1}-n=\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k} \Rightarrow \\
\color{green}{\frac{x(x+1)^{n-1}}{(x+1)^n-1}-1=\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}}$$
Altogether 
$$\color{red}{\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k+1}{x-x_k}}=
\color{green}{\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{x_k}{x-x_k}}+
\frac{1}{n}\color{blue}{\sum\limits_{k=0}^{n-1}\frac{1}{x-x_k}}=\\
\color{green}{\frac{x(x+1)^{n-1}}{(x+1)^n-1}-1} + \frac{1}{n}\color{blue}{\frac{n(x+1)^{n-1}}{(x+1)^n-1}}=\\
\frac{(x+1)^{n}}{(x+1)^n-1}-1=
\color{red}{\frac{1}{(x+1)^n-1}}$$
What's left now is to show that the roots of $P(x)$ are $$x_k=a(k,n)-1=e^{\frac{2 k\pi}{n}\cdot i}-1, \space k=\overline{0..n-1}$$
which is an easy exercise. You will see similar techniques being used here, here and here.
