<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] Here is a more "1st principles" pf. I use a hint in Marsden's book.
1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas.

Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy 
$$(1-z)^n=1 \leftrightarrow (1-z) \in \left\{\cos \frac{2 \pi k}{n}+i \sin \frac{2\pi k}{n}:k=0,...,n-1 \right\}$$
(the set of nth roots of 1) 
$$\leftrightarrow z \in \left\{z_k= 1-\cos \frac{2 \pi k}{n}-i \sin \frac{2\pi k}{n}:k=0,...,n-1\right\}\;\;\; (2)$$.

Since $z_0,....,z_{n-1}$ are the roots of $(1-z)^n-1$, we have by factorization that 
$$(1-z)^n-1=\prod_{k=0}^{n-1}(z_k-z)=-z \prod_{k=1}^{n-1}(z_k-z) \;\;(3)$$ (since, by (2), $z_0=0$)

In (3), the LHS and RHS are polynomials in z. Equating the coeffs in front of z, we get 
$$-n=-\prod_{k=1}^{n-1}z_k \leftrightarrow n=\prod_{k=1}^{n-1}z_k\,.$$  
Note 
$$\prod_{k=1}^{n-1} \bar{z}_k=\overline{\prod_{k=1}^{n-1}z_k}=n$$
(since $n\in \mathbb{R}$), so 
$$\prod_{k=1}^{n-1}|z_k|^2=\prod_{k=1}^{n-1} z_k \bar{z}_k=\prod_{k=1}^{n-1} z_k \prod_{k=1}^{n-1} \bar{z}_k=n^2\;\; (4).$$

Next, 
$$|z_k|^2=(1-\cos \frac{2 \pi k}{n})^2+ \sin^2 \frac{2\pi k}{n}=2(1-\cos \frac{2 \pi k}{n})$$
using this in (4) gives 
$$2^{n-1} \prod_{k=1}^{n-1}(1-\cos \frac{2 \pi k}{n})=n^2\;\;(5)$$.

Next, 
$$(\prod_{k=1}^{n-1} \sin \frac{k \pi}{n})^2=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \prod_{k=1}^{n-1} \sin \frac{(n-k) \pi}{n}=\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} \sin \frac{(n-k) \pi}{n}=$$
(where in the last 2 steps, we exploit that the order of taking a product doesn't matter)
$$=\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (\cos \frac{(n-2k) \pi}{n}-\cos \pi)=$$
(by (1))
$$=\frac{1}{2^{n-1}} \prod_{k=1}^{n-1} (1-\cos \frac{2k \pi}{n})=$$ 
(using $\cos (\pi -x)=-\cos x$) 
$$=n^2 /2^{2(n-1)}\;.$$ Applying a sqrt to everything gives the desired result.

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