<s>[INST] 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.
 [/INST] Clearly $z=0$ is a root of $(z+1)^n-1$, so we need to exclude it to get something nontrivial for the product. Dividing and taking the limit,
$$ \lim_{z \to 0} \frac{(z+1)^n-1}{z} = n, $$
so the product of the remaining roots is $(-1)^{n-1} n$. Now we have to find expressions for the roots. We have roots
$$ z_k+1 = e^{2\pi ik/n}, \quad k \in \{1,2,\dotsc,n-1\}. $$
Therefore, rearranging and applying the formula for sine,
$$ z_k = e^{2\pi i k/n} -1 = e^{\pi i k/n} (e^{\pi i k/n}-e^{-\pi ik/n}) = e^{\pi i k/n} 2i\sin{\left( \frac{\pi k}{n} \right)}. $$
Hence, we have
$$ \begin{align}
(-1)^{n-1}n &= \prod_{k=1}^{n-1} z_k \\
&= \prod_{k=1}^{n-1} e^{\pi i( k/n+1/2)} 2\sin{\left( \frac{\pi k}{n} \right)} \\
&= 2^{n-1} \exp{\left( \pi i \left(\frac{n-1}{2}+\frac{1}{n} \sum_{k=1}^{n-1} k \right) \right)}  \prod_{k=1}^{n-1} \sin{\left( \frac{\pi k}{n} \right)}.
\end{align} $$
The result now follows, since
$$ \exp{\left( \pi i \left(\frac{n-1}{2}+\frac{1}{n} \sum_{k=1}^{n-1} k\right) \right)} = \exp{\left( \pi i (n-1) \right)} = (-1)^{n-1}. $$
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