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.
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.
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.
