Let $z_k$ be a $n$-th root of the unity, i.e.
$$z_k = e^{2\pi i\frac{k}{n}}\\ k\in \{0,1,\cdots,n-1\}$$
Prove that $$(z_0 - z_1)(z_0 - z_2)\ldots (z_0 - z_{n-1})= n$$
This problem was at the end of the introduction chapter of a book about complex numbers, way before talking about poles and residues. I thought of doing it by induction but for $n+1$ we don't have the same $z_k$ as for $n$.
What surprises me the most is the fact that $z_k = \bar{z}_{n-k}$ So:
$$(z_0 - z_k)(z_0 - z_{n-k}) = 2 - 2\Re(z_k)$$
Thus the total product becomes:
$$2^{\lceil\frac{n}{2}\rceil}\prod_{k=1}^{\lceil\frac{n}{2}\rceil}(1-\Re(z_k))\stackrel{?}{=}n$$ Any hints?
We know that $$ (x-1)(x-z_1)\ldots(x-z_{n-1})=x^n-1. $$ Hence $$ (x-z_1)\ldots(x-z_{n-1})=\frac{x^n-1}{x-1}=x^{n-1}+x^{n-2}+\ldots+x+1. $$ If take $x=1$, then $$ (1-z_1)\ldots(1-z_{n-1})=n. $$