How to prove that $2^{n-1} \prod_{k=1}^{n-1} (1- \cos(\frac{2 \pi}{n} k)) = n^2$?

Question

I found numerically the interesting formula that for $n \in \mathbb{N}$, $$2^{n-1} \prod_{k=1}^{n-1}\left(1- \cos\left(\frac{2 \pi}{n} k\right)\right) = n^2$$

I've been struggling to find a proof. For context, this is the product of nonzero eigenvalues of the Laplacian matrix of a circle consisting of $n$ points.

Does anyone know a way to prove this?

Answer 1

Let $\zeta=\exp(2\pi i/n)$. Then this product is $$P=\prod_{k=1}^{n-1}\left(1-\frac{\zeta^k+\zeta^{-k}}{2}\right) =\frac1{2^{n-1}}\prod_{k=1}^{n-1}(1-\zeta^k)(1-\zeta^{-k}).$$ Therefore $P=2^{-n+1}A^2$ where $$A=\prod_{k=1}^{n-1}(1-\zeta^k).$$ This is well-known to equal $n$. For a proof, consider the factorisation $$x^n-1=\prod_{k=0}^{n-1}(x-\zeta^k)$$ and divide by $x-1$ to get $$x^{n-1}+x^{n-2}+\cdots+x+1=\prod_{k=1}^{n-1}(x-\zeta^k).$$ Set $x=1$ here to give $n=A$