I'm in a question about polygonals and got stuck at a part. I have to prove that
$$\prod_{k=1}^{n-1} \left(1 - \operatorname{cis}(\frac{2k\pi}{n})\right) = n$$
I've tried to multiply it to make $\operatorname{cis}(\frac{2k\pi}{n})$ transform to $\operatorname{cis}(\frac{2k\pi}{n})^n=1$, but it doesn't help.
The factors $1 - \mathop{\text{cis}} \frac{2k\pi}{n}$, $k = 1,\ldots, n - 1$, are the distinct roots of the polynomial
$$(1 - x)^{n - 1} + (1 - x)^{n - 2} + \cdots + (1 - x) + 1.$$
The product of the roots of any polynomial is its constant term, which is $n$ in this case.
In case the first statement is not clear: the roots of $y^n - 1$ are $\mathop{\text{cis}} \frac{2k\pi}{n}$ for $k = 0,\ldots, n-1$, so the roots of $$\frac{y^n - 1}{y - 1} = y^{n - 1} + y ^{n - 2} + \cdots + y + 1$$ are $\mathop{\text{cis}} \frac{2k\pi}{n}$ for $k = 1,\ldots, n-1$. Set $y = 1 - x$.
Write $$ \prod_{k=1}^{n-1} \left(x - \operatorname{cis}(\frac{2k\pi}{n})\right) = \frac{x^n - 1}{x - 1} = x^{n - 1} + x ^{n - 2} + \cdots + x + 1 $$ Now set $x=1$.
