How to prove that :$\prod_{k=0}^{n-1}e^{\frac{2\pi i k}{n}}=(-1)^{n-1}\;\;\; n\in\mathbb{N}^*$

Question

Can someone tell me how to prove the folowing equalty : $$\prod_{k=0}^{n-1}e^{\frac{2\pi i k}{n}}=(-1)^{n-1}\;\;\; n\in\mathbb{N}^*.$$

Thanks in advance.

Answer 1

I will answer my question

We have : $$\prod_{k=0}^{n-1}e^{\frac{2\pi i k}{n}}=e^{\frac{2\pi i }{n}\frac{n(n-1)}{2}}=e^{\pi i (n-1)}=\cos((n-1)\pi)=(-1)^{n-1}.$$

Answer 2

Consider the polynomial $x^n-1$. Vieta's formula says that $-1=(-1)^n z_1\cdots z_n$ where the $z_i$ are the roots of the polynomial. What are these roots?