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

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 $$ \exp{\left( \pi i \left(\frac{n-1}{2}+\frac{1}{n} \sum_{k=1}^{n-1} k\right) \right)} = \exp{\left( \pi i (n-1) \right)} = (-1)^{n-1}. $$

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, $$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}(e^{\pi i \frac{k}{n}} - e^{-\pi i \frac{k}{n}}) = (-1)^{n-1}n$$ $$\prod_{k = 1}^{n-1} e^{\pi i \frac{k}{n}}2i \sin \frac{\pi k}{n} = (-1)^{n-1}n$$ $$e^{\pi i \frac{n-1}{2}}2^{n-1}i^{n-1}\prod_{k = 1}^{n-1} \sin \frac{\pi k}{n} = (-1)^{n-1}n$$ $$\prod_{k = 1}^{n-1}\sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}$$