TITLE: Proving $(1-x)\cdot (1-x^2)\cdots(1-x^{n-1})=n$ if $x^n=1$ and $x\neq 1$
QUESTION [5 upvotes]: If we have a equation $x^n=1$, then how can we prove
$$(1-x)\cdot (1-x^2)\cdots (1-x^{n-1})=n $$
when $x$ is not $1$? I know that $x= e^{(2\pi + 2k\pi)/n}$ and we can get different value of $x$ when $ k=0,1....,n-1$ but expanding our product will be a heavy task, is there any simpler way around?

REPLY [2 votes]: If $\xi$ is any primitive $n$-th root of unity, we have:
$$ \frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}\left(z-\xi^{k}\right)=\prod_{k=1}^{n-1}(1-z \xi^k) \tag{1}$$
and the (right) claim follows from De L'Hopital theorem, by computing $\lim_{z\to 1^-}$ of both sides.