IRL_ABooklet

Note that for any complex number $\zeta$ with $|\zeta |=1$ we have $$|1-\zeta {|}^2=(1-\zeta )(1-\bar{\zeta })=2-(\zeta +\bar{\zeta })=2(1-\mathrm{Re}(\zeta )).$$ If $\zeta =e^{2i\theta }=\cos (2\theta )+i\sin (2\theta )$, we can use $1-\cos (2\theta )=2{\sin }^2\theta$ to obtain $$|1-e^{2i\theta }{|}^2=4{\sin }^2\theta .$$ Therefore, $$4^{n-1}{\left(\prod _{k=1}^{n-1}\sin \frac{k\pi }{n}\right)}^2=\prod _{k=1}^{n-1}\left(4{\sin }^2\frac{k\pi }{n}\right)=\prod _{k=1}^{n-1}{\left|1-e^{\frac{2ik\pi }{n}}\right|}^2.$$ Now, the $n$ numbers $e^{\frac{2ik\pi }{n}},k=0,1,\dots ,n-1$ are precisely the $n$th roots of unity, and so $$z^n-1=(z-1)\prod _{k=1}^{n-1}\left(z-e^{\frac{2ik\pi }{n}}\right).$$ Hence, for all $z$, $$z^{n-1}+\cdots +z+1=\prod _{k=1}^{n-1}\left(z-e^{\frac{2ik\pi }{n}}\right),$$ and so, letting $z=1$, $$n=\prod _{k=1}^{n-1}\left(1-e^{\frac{2ik\pi }{n}}\right),\text{which gives}{n}^2=\prod _{k=1}^{n-1}{\left|1-e^{\frac{2ik\pi }{n}}\right|}^2.$$ Consequently, $$2^{2(n-1)}{\left(\prod _{k=1}^{n-1}\sin \frac{k\pi }{n}\right)}^2=n^2,$$ from which it follows that $$\prod _{k=1}^{n-1}\sin \frac{k\pi }{n}=\frac{n}{2^{n-1}},$$ since $\sin \frac{k\pi }{n}>0$ for $k=1,2,\dots ,n-1$.