Very complex Calgebra

\begin{aligned} 2^{n-1}\prod_{k=1}^{n-1}\left(\cos(\theta)-\cos\left(\dfrac{k\pi}n\right)\right) &=\lim_{\theta \rightarrow 0}\prod_{k=1}^{n-1}\left(u+\frac1u-e^{ik\pi/n}-e^{-ik\pi/n}\right)\tag{1}\\ &=\lim_{\theta \rightarrow 0}\frac1{u^{n-1}}\prod_{k=1}^{n-1}\left(u-e^{ik\pi/n}\right)\prod_{k=1}^{n-1}\left(u-e^{-ik\pi/n}\right)\quad\quad\quad\tag{2}\\ &=\lim_{\theta \rightarrow 0}\dfrac1{u^{n-1}}\dfrac{u^{2n}-1}{u^2-1}\tag{3}\\[3pt] &=\lim_{\theta \rightarrow 0}\dfrac{u^n-u^{-n}}{u-u^{-1}}\tag{4}\\[6pt] &=\lim_{\theta \rightarrow 0}\dfrac{\sin(n\theta)}{\sin(\theta)}=\dfrac{n\theta\cdot\dfrac{\sin(n\theta)}{n\theta}}{\theta\cdot\dfrac{\sin(\theta)}{\theta}}=\boxed{n}\tag{5} \end{aligned}

Explanation:

(1): distribute $2$ over each term in the product

(2): $u+\dfrac1u-e^{ik\pi/n}-e^{-ik\pi/n}=\dfrac1u\left(u-e^{ik\pi/n}\right)\left(u-e^{-ik\pi/n}\right)$

(3): $\prod\limits_{k=1}^{2n}\left(u-e^{ik\pi/n}\right)=u^{2n}-1$ , then divide out the $k=n$ and $k=2n$ terms

(4): algebra

(5): $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and $\displaystyle\lim_{x \rightarrow 0}\dfrac{sinx}{x}=1$