<s>[INST] Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that 
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
 [/INST] Consider $z^n=1$, each root is 
$$\xi_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} = e^{i\frac{2k\pi}{n}}, k=0,1,2,...,n-1 $$
So, we have
$$ z^n -1 = \prod_{k=0}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-\xi_0)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-1)\prod_{k=1}^{n-1}(z-\xi_k)$$
$$\Longrightarrow z^{n-1}+...+z^2+z+1 = \prod_{k=1}^{n-1}(z-\xi_k)$$
By  substituting  z=1, $$\Longrightarrow n = \prod_{k=1}^{n-1}(1-\xi_k) $$
Next, take the modulus on both sides,
$$ |n| = n = |\prod_{k=1}^{n-1}(1-\xi_k)| = \prod_{k=1}^{n-1}|(1-\xi_k)|$$
$$ 1 - \xi_k = 1-(\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}) = 2\sin\frac{k\pi}{n}(\sin\frac{k\pi}{n} -i\cos\frac{k\pi}{n})$$
$$ |1 - \xi_k| = 2\sin\frac{k\pi}{n} $$
So, 
$$ n = 2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}$$
$$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} $$
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