Using $\text{n}^{\text{th}}$ root of unity
$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$
Prove that
$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
Using $\text{n}^{\text{th}}$ root of unity
$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$
Prove that
$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$ $(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$ where $\xi=e^{2i\pi/n}$. Now note, that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, thus cancelling $x-1$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$. Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. Therefore $P=n2^{1-n}$.
Edit:
In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by polynomial reminder theorem we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees of L.H.S. and R.H.S. we can find, that $Q(x)$ has degree $0$. Comparing highest coefficients we can conclude $Q(x)=1$.
Edit: We may instead use the identity $\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2\sin(k\pi/n), k = 1, ..., n - 1,$ to establish immediately that $P \equiv \prod_{k=1}^{n-1}\sin(k\pi/n)= 2^{1-n}\prod_{k=1}^{n-1}\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2^{1 - n}\left\lvert \prod_{k=1}^{n-1}(1 - e^{2ik\pi/n}) \right\rvert$, and continue by applying the foregoing logic to the product to obtain $P=n2^{1-n}$.
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}} $
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.