How find this value of $\prod_{1\le i<j\le n}(w^i-w^j)^2$

Question

give the positive integer number $n$, and $w=\cos{\dfrac{2\pi}{n}}+i\sin{\dfrac{2\pi}{n}}$ where $i^2=-1$

find the vaule

$$\prod_{1\le i<j\le n}(w^i-w^j)^2$$

My try:note

$$w^n=1$$ $$\prod_{1\le i<j\le n}(w^i-w^j)^2=\prod_{1\le i<j\le n}(w^i-w^j)(w^i-w^j)$$

and I know this $$\prod_{i=1}^{n-1}(1-w^i)=n$$ becasuse $$1+x+x^2+\cdots+x^{n-1}=(x-w)(x-w^2)\cdots (x-w^{n-1})$$ $$\prod_{1\le i<j\le n}(w^i-w^j)^2=\prod_{1\le i<j\le n}w^{2i}(1-w^{j-i})^{2}$$ Then I can't works,Thank you

Answer 1

Hint: First, find out the value of $$\prod_{i=1}^{n-1} (1-w^i)$$