<s>[INST] Using complex numbers prove that $\sin(\frac {\pi}{m})\sin(\frac {2\pi}{m})\ldots\sin(\frac {(m-1)\pi}{m})= \frac{m}{2^{m -1}}$ Prove that for $m=2,3,\dots$
$$
\sin\left(\frac{\vphantom{1}\pi}m\right)\sin\left(\frac{2\pi}m\right)\sin\left(\frac{3\pi}m\right)\cdots\sin\left(\frac{(m-1)\pi}m\right)=\frac{m}{2^{m-1}}
$$
I have no idea how to begin at all, I'm trying to think of a way using de Moivre's theorem but I can't seem to figure it out.
 I'm sorry for not showing effort but I'm completely stuck.
 [/INST] $|\cos \theta + i\sin \theta - 1|\\
\sqrt {1 - 2\cos \theta} = 2\sin \frac {\theta}{2}\\
2\sin\theta = |e^{2\theta i} - 1|$
$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1|$ 
The set $\{e^{\frac{2n\pi}{m} i}\}$ make up the roots of $z^m - 1 = 0$ excluding the root at $z= 1$ or the roots of $(z^{m-1} + z^{m-2} + z^{m-3} + \cdots  + 1)$
$(z - e^{\frac{2\pi}{m}i})(z - e^{\frac{4\pi}{m}i})(z - e^{\frac{6\pi}{m}i})\cdots(z - e^{\frac{(m-2)\pi}{m}i})= (z^{m-1} + z^{m-2} + z^{m-3} + \cdots  + 1)\\
|z - e^{\frac{2\pi}{m}i}||z - e^{\frac{4\pi}{m}i}||z - e^{\frac{6\pi}{m}i}|\cdots|z - e^{\frac{(m-2)\pi}{m}i}|= |z^{m-1} + z^{m-2} + z^{m-3} + \cdots  + 1| $
and now set z = 1
$|1 - e^{\frac{2\pi}{m}i}||1 - e^{\frac{4\pi}{m}i}||1 - e^{\frac{6\pi}{m}i}|\cdots|1 - e^{\frac{(m-2)\pi}{m}i}| = |1^{m-1} + 1^{m-2} + 1^{m-3} + \cdots  + 1| = m$
$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1| = \frac {m}{2^{m-1}}$ 
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