$z_1,z_2,…z_n$ satisfy the equation $(z+1)^n=1$. Use the product of $z_1,z_2,…z_n$ to prove that $$\sin \frac {\pi}{n} \sin \frac {2\pi}{n}…\sin \frac {(n-1)\pi}{n}=\frac {n}{2^{n-1}}$$
Attempt I compute $z_1=1-1=0,…z_n=\cos \frac{(n-1)2\pi }{n}+i\sin \frac{(2n-1)\pi }{n}-1$
But if I do the multiplication ,the calculation result seems very ugly...
Hint. Note that $$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\prod_{k=1}^{n-1}\frac{e^{i\frac{k\pi}{n}}-e^{-i\frac{k\pi}{n}}}{2i}= \frac{e^{-i\frac{n(n-1)\pi}{2n}}}{(2i)^{n-1}}\prod_{k=1}^{n-1}\left(e^{i\frac{2k\pi}{n}}-1\right) =\frac{(-1)^{n-1}\prod_{k=2}^{n}z_{k}}{2^{n-1}}.$$ where $$\frac{(z+1)^n-1}{z}=z^{n-1}+nz^{n-2}+\dots +n=(z-z_2)\dots(z-z_n).$$ Can you take it from here?
Consider using the Viete's theorem.
We need to factor out your first root $z=0$. After that the constant term is $n$, and the highest term is $z^{n-1}$. So the product of the roots should be $(-1)^{n-1}n$.
But I can't relate the product to the main problem yet...
A different approach. Consider $x^{2n}-2x^n\cos(na)+1=0 ,\ n\gt0 \, , a\gt 0$. This gives you $x^n=\exp \pm i na$. or $x=\exp \pm i\left(a+\frac{2\pi k}{n}\right), \ k=0, 1, 2,\ldots , n-1$.
$$\begin{aligned}x^{2n}-2x^n\cos(na)+1&=0\\ \prod_{k=0}^{n-1}\left(x-\exp i\left(\dfrac{a+2\pi k}{n}\right)\right)\left(x-\exp i\left(-\dfrac{a+2\pi k}{n}\right)\right)&=0\\ \prod_{k=0}^{n-1}\left[ x^2-2x\cos\left(a+\dfrac{2\pi k}{n}\right)+1\right]&=0\end{aligned}$$
Let $x=1$ and set $a=2b$. $$\begin{aligned}\prod_{k=0}^{n-1}\left[1-2\cos\left(2b+\dfrac{2\pi k}{n}\right)+1\right]&=0\\ \prod_{k=0}^{n-1}\left[2-2\cos2\left(b+\dfrac{\pi k}{n}\right)\right]&=0\end{aligned}$$
$$\begin{aligned}\prod_{k=0}^{n-1}\left[2-2\cos2\left(b+\dfrac{k\pi}{n}\right)\right]&=2-2\cos\left(2nb\right)\equiv4\sin^2\left(nb\right)\end{aligned}\tag1$$
Also $$\begin{aligned}2-2\cos2\left(b+\dfrac{\pi k}{n}\right)&=2-2\left[\cos^2\left(b+\dfrac{\pi k}{n}\right)-\sin^2\left(b+\dfrac{\pi k}{n}\right)\right]\\&=4\sin^2\left(b+\dfrac{k\pi }{n}\right)\end{aligned}\tag2$$
Equate $(1)$ and $\prod_{k=0}^{n-1}(2)$ and take square root of both sides to get the desired result.
$$\begin{aligned}\sqrt{4\sin^2\left(nb\right)}&=\sqrt{\prod_{k=0}^{n-1}4\sin^2\left(b+\dfrac{k\pi }{n}\right)}\\2\sin \left(nb\right)&=2^n\prod_{k=0}^{n-1}\sin\left(b+\dfrac{k\pi }{n}\right)\\ \dfrac{\sin\left(nb\right)}{\sin b}&=2^{n-1}\dfrac{\sin\left(b\right)}{\sin b}\sin\left(b+\dfrac{\pi}{n}\right)\cdots\sin\left(b+\dfrac{(n-1)\pi}{n}\right)\end{aligned}$$
Now take the limit as $b\to 0^+$ which gives you: $$n=2^{n-1}\underbrace{\sin\left(\dfrac{\pi}{n}\right)\sin\left(\dfrac{2\pi}{n}\right)\cdots\sin\left(\dfrac{(n-1)\pi}{n}\right)}_{\text{Let it }=\varphi}\implies \boxed{\varphi =\dfrac{n}{2^{n-1}}}$$
Consider the expression $$(z+1) = Z $$ Then, $$Z^{n}=1 $$ Also, $$1=(-1)^{2}=e^{i2k\pi} $$ Therefore, $$Z=(1)^{1/n}=e^{i2k\pi/n} $$ And then, $$ z=e^{i2k\pi/n}-1$$ Now, this has got $n$ roots for $ k =0 \space to \space n-1$ and we designate these roots as $z_0,z_1,z_2...z_{n-1}$. Now, coming back to original expression and expanding it: $$\begin{align*} (z+1)^{n}&=1\\ z^{n}+nz^{n-1}+\frac n2(n-1)z^{n-2}+..+nz+1&=1\\ z^{n}+nz^{n-1}+\frac n2(n-1)z^{n-2}+..+nz&=0 \end{align*}$$
$z=0$ is a root. So divide above equation by z and you will get: $$\begin{align*} z^{n-1}+nz^{n-2}+\frac n2(n-1)z^{n-3}+..+n&=0 \end{align*}$$ The above equation now has roots $z_1,z_2,...,z_{n-1}$ and the product of these roots will be $(-1)^{n-1}n$ (ratio of coefficients of lowest power term to highest power term, positive for even degree polynomial and negative for odd degree polynomial). $$\prod_{k=1}^{n-1}\left(e^{i\frac{2k\pi}{n}}-1\right)=(-1)^{n-1}n $$
Now using the hint by Robert Z:
$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{1}{(2i)^{n-1}}\prod_{k=1}^{n-1}\left(e^{i\frac{k\pi}{n}}-e^{-i\frac{k\pi}{n}}\right)= \frac{1}{(2i)^{n-1}}\prod_{k=1}^{n-1}e^{-i\frac{k\pi}{n}}\left(e^{i\frac{2k\pi}{n}}-1\right).$$
And, $$\frac{1}{(2i)^{n-1}}\prod_{k=1}^{n-1}e^{-i\frac{k\pi}{n}}\left(e^{i\frac{2k\pi}{n}}-1\right) = \frac{(-1)^{n-1}n}{(2i)^{n-1}}\prod_{k=1}^{n-1}e^{-i\frac{k\pi}{n}} = \frac{(-1)^{n-1}n}{(2i)^{n-1}}e^{-i\frac{(n-1)\pi}{2}}$$
Which on simplification, gives $$\frac{(-1)^{n-1}n}{(-2)^{n-1}} = \frac{n}{(2)^{n-1}}$$
