A product involving complex numbers

Question

Given $ω^3=1$ but $ω≠1$ how would you prove that, $\displaystyle \prod_{m=1}^{2n} (1-ω^{2^m}+ω^{2^{m+1}})=2^{2n}$ ?

I tried but made no progress tackling this.

Answer 1

$\displaystyle \prod_{m=1}^{2n} (1-ω^{2^m}+ω^{2^{m+1}})=(1-ω^{2}+ω^{4})*(1-ω^{4}+ω^{8})*(1-ω^{8}+ω^{16})*.... 2n \ times.$

We know $ω^{3k+p}=ω^{p}$

$=(1+ω-ω^{2})*(1-ω+ω^{2})*(1+ω-ω^{2})*.... 2n \ times.$

$= (1-ω+ω^{2})*(1-ω+ω^{2})*...\ n \ times * (1+ω-ω^{2})*(1+ω-ω^{2})*... n \ times.$

$=(1-ω+ω^{2})^{n}*(1+ω-ω^{2})^{n}$

$=(1-ω+ω^{2}+(ω-ω))^{n}*(1+ω-ω^{2}+(ω^{2}-ω^{2}))^{n}$

As we know $1+ω+ω^{2}=0$ (Ref - Cube roots of Unity) , we get :

$=(-2ω^{2})^{n}*(-2ω)^{n}$

$= (4ω^{3})^{n}=4^{n}=2^{2n}$