TITLE: Product Identity Multiple Angle or $\sin(nx)=2^{n-1}\prod_{k=0}^{n-1}\sin\left(\frac{k\pi}n+x\right)$
QUESTION [6 upvotes]: I have run across this interesting identity that I am unable to verify.
$$\sin(nx)=2^{n-1}\prod_{k=0}^{n-1}\sin\left(\frac{k\pi}n+x\right)$$
Can anyone provide a hint as how one would prove this? 
This identity was found on the Wolfram site1.

REPLY [10 votes]: This solution herein is inspired by a comment left by @user1952009 on another posted solution.  That solution is elegant in that it relies on only $(i)$ Euler's formula to write 
$$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}\tag 1$$ 
and $(ii)$ the $n$ roots of unity, $z=e^{-i2k \pi/n}$ for $k=0,\dots,n-1$, of the equation $z^n=1$ to write 
$$z^n-1=\prod_{k=0}^{n-1}(z-e^{-i2\pi k/n})\tag 2$$
Proceeding, we find  
$$\begin{align}
\sin(nx)&=\frac{e^{inx}-e^{-inx}}{2i}&\\\\
&=\frac{e^{-inx}}{2i}\left(e^{i2nx}-1\right)\\\\
&=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}(e^{i2x}-e^{-i2\pi k/n})\\\\
&=\frac{e^{-inx}}{2i}\prod_{k=0}^{n-1}\left(\color{blue}{e^{ix}}\color{red}{e^{-i\pi k/n}}\color{green}{(2i)}\sin\left(x+\frac{k \pi}{n}\right)\right)\\\\
&=\frac{e^{-inx}}{2i} \color{blue}{e^{inx}} \color{red}{e^{-i(\pi/n)n(n-1)/2}}\color{green}{(2i)^n}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)\\\\
&=2^{n-1}\prod_{k=0}^{n-1}\sin\left(x+\frac{k \pi}{n}\right)
\end{align}$$
as was to be shown!