How to prove $ \prod_{k=1}^{n-1}\left(e^{i\frac{2k\pi}{n}}-1 \right) = (-1)^{n-1} n $

Question

how to prove the following equation? $$ \prod_{k=1}^{n-1}\left(e^{i\frac{2k\pi}{n}}-1 \right) = (-1)^{n-1} n $$

The product on the LHS is hard to calculate. The only progress I got is that each term on the LHS is the root of equation $(z+1)^n=1$.

Thanks!