prove $\frac{1}{x^n+1}=\prod_{k=1}^n(x-x_k)^{-1}$

Question

How it can shown that:

$$\frac{1}{x^n+1}=\prod_{k=1}^n(x-x_k)^{-1}$$

where $x_k=e^{i(2k-1)\pi/n}$, $k=1, \cdots,n$

I just know that: $$\frac{1}{x^{n}+1}=\left(x^{n}+1\right)^{-1}=\sum_{k=0}^{\infty}\binom {-1}{k}x^{nk}$$

Using negative binomial theorem it can be written as:

$$\sum_{k=0}^{\infty}(-1)^kx^{nk}$$

Another representation for $\frac{1}{x^n+1}$ is:

$$\frac{1}{x^n+1}=\sum_{k=1}^na_k(x-x_k)^{-1} $$

where $a_k=\frac{-x_k}{n}$

But I don't know where does that come from.

Answer 1

$x_k^n=-1$ implies that $x^n+1=(x-x_1)....(x-x_n)$ since $x_1,...,x_k$ are roots of $x^n+1$.

Answer 2

Your initial steps are sufficient. $$x^n+1=(x-x_1)(x-x_2)(x-x_3)..(x-x_k)...(x-x_n)=\prod_{k=1}^{k=n} (x-x_k)$$ where $x_k=e^{i(2k-1)\pi/n},k=1,2,3,...n.$ Hence the result by inverting both sides.

The other result is due to Partial Fractions: $$\frac{P(x)}{Q(x)}=\sum_{k=1}^{n} \frac{P(x_k)}{Q'(x_k)} \frac{1}{x-x_k}=\sum_{k=1}^{n} \frac{A_k}{x-x_k}.$$ where $P(x)$ and $Q(x)$ are polynomials, former is of lesser degree than that of $Q(x)$ and $Q(x_k)=0.$ Here $P(x)=1$ and $Q(x)=x^n+1$. So $A_k=\frac{1}{n x_k^{n-1}}=\frac{-x_k}{n}$ as $x_k^n=-1$ You may see: https://en.wikipedia.org/wiki/Partial_fraction_decomposition