$\sin(\frac{\pi}{n})\sin(\frac{2\pi}{n})...\sin(\frac{(n-1)\pi}{n})=\frac{n}{2^{n-1}}$ Prove that $$\sin\left(\frac{\pi}{n}\right)\sin\left(\frac{2\pi}{n}\right)\sin\left(\frac{3\pi}{n}\right).....\sin\left(\frac{(n-1)\pi}{n}\right)=\frac{n}{2^{n-1}}$$ 
Is there a proof without using complex numbers and $n-th$ roots of unity.
 A: The following is the simplest proof I know. We have the identity
\begin{equation} 
x^{2n} - 2x^n y^n \cos n\theta + y^{2n} = \bigg\{x^2 -2xy \cos \theta + y^2\bigg\}\bigg\{x^2-2xy \cos \bigg(\theta+\frac{2\pi}{n}\bigg)+y^2\bigg\}\cdots
\end{equation} 
to $n$ factors adding $2\pi/n$ to each angle successively.  This can be seen by noting the LHS and RHS share the same roots in $x$ using complex numbers, but given complex numbers are a trigonometric convenience I imagine there's a non-complex way to arrive at the identity. Let $x=y=1$, $\theta = 2\phi$, and apply $1-\cos\theta = 2\sin(\theta/2)$
\begin{equation}
\sin n\phi = 2^{n-1} \sin \phi \sin\bigg(\phi + \frac{\pi}{n}\bigg)\sin\bigg(\phi + \frac{2\pi}{n}\bigg) \cdots \sin\bigg(\phi + \frac{(n-1)\pi}{n}\bigg)
\end{equation}
Divide by $\sin \phi$ and let $\phi \rightarrow 0$ to get the equation 
A: Hint:
Observe that we have
$$
z^{n-1}+\cdots+z+1=\bigg(z-\exp(\frac{2\pi i}n)\bigg)\cdots\bigg(z-\exp(\frac{2\pi i(n-1)}n)\bigg)
$$
Now let $z=1$ and use some trigonometry at the righthand side...
