Prove that $\sin\frac{\pi}n·\sin\frac{2\pi}n···\sin\frac{(n-1)\pi}n=\frac{n}{2^{n-1}}$ How to prove that

$$
\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}
$$

using the roots of $(z+1)^n-1=0$?
My rough idea is to solve  $(z+1)^n-1=0$ and use De Moivre's Theorem to find the product of roots to prove the equality.

Given the equation
$$(z+1)^n = 1 \tag{1}$$
Its roots are, $$z = e^{2\pi i \frac{k}{n}} - 1, \;\; 0 \leq k < n$$
Now,
$$\prod_{k = 1}^{n-1} (e^{2\pi i \frac{k}{n}} - 1)$$
is the product of all roots except $z = 0$. If you open up (1) and cancel one out of from both sides, you can take z common and the remaining polynomial will have the other roots. From Vieta's formulae, we know that,
$$\prod_{k = 1}^{n-1} (e^{2\pi i \frac{k}{n}} - 1) = (-1)^{n-1}n$$
The above can be written as,
$$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}(e^{\pi i \frac{k}{n}} - e^{-\pi i \frac{k}{n}}) = (-1)^{n-1}n$$
$$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}2i \sin \frac{\pi k}{n} = (-1)^{n-1}n$$
$$e^{\pi i \frac{n-1}{2}}2^{n-1}i^{n-1}\prod_{k = 1}^{n-1} \sin \frac{\pi k}{n} = (-1)^{n-1}n$$
$$\prod_{k = 1}^{n-1}\sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}$$
