Note that for any complex number $\zeta$ with $|\zeta| = 1$ we have
$$|1 - \zeta|^2 = (1 - \zeta)(1 - \bar{\zeta}) = 2 - (\zeta + \bar{\zeta}) = 2(1 - \operatorname{Re}(\zeta)).$$
If $\zeta = e^{2i\theta} = \cos(2\theta) + i\sin(2\theta)$, we can use $1 - \cos(2\theta) = 2\sin^2\theta$ to obtain
$$|1 - e^{2i\theta}|^2 = 4 \sin^2 \theta.$$
Therefore,
$$4^{n-1} \left( \prod_{k=1}^{n-1} \sin \frac{k\pi}{n} \right)^2 = \prod_{k=1}^{n-1} \left( 4 \sin^2 \frac{k\pi}{n} \right) = \prod_{k=1}^{n-1} \left| 1 - e^{\frac{2ik\pi}{n}} \right|^2.$$
Now, the $n$ numbers $e^{\frac{2ik\pi}{n}}, k = 0, 1, \dots, n-1$ are precisely the $n$th roots of unity, and so
$$z^n - 1 = (z - 1) \prod_{k=1}^{n-1} \left(z - e^{\frac{2ik\pi}{n}}\right).$$
Hence, for all $z$,
$$z^{n-1} + \dots + z + 1 = \prod_{k=1}^{n-1} \left(z - e^{\frac{2ik\pi}{n}}\right),$$
and so, letting $z = 1$,
$$n = \prod_{k=1}^{n-1} \left(1 - e^{\frac{2ik\pi}{n}}\right), \quad \text{which gives} \quad n^2 = \prod_{k=1}^{n-1} \left|1 - e^{\frac{2ik\pi}{n}}\right|^2.$$
Consequently,
$$2^{2(n-1)} \left( \prod_{k=1}^{n-1} \sin \frac{k\pi}{n} \right)^2 = n^2,$$
from which it follows that
$$\prod_{k=1}^{n-1} \sin \frac{k\pi}{n} = \frac{n}{2^{n-1}},$$
since $\sin \frac{k\pi}{n} > 0$ for $k = 1, 2, \dots, n-1$.