TITLE: a problem on a special root of $x^{11}-1=0$
QUESTION [6 upvotes]: I have came across the following problem
if $\alpha$ be a special root of the equation $x^{11}-1=0$ , then prove that  $$(\alpha+1)(\alpha^2+1)......(\alpha^{10}+1)=1$$

totally stuck on it. how to solve this.please help me somebody.

REPLY [12 votes]: If $\alpha$ is a 'special root', in this case meaning that $\alpha\ne 1$, then using $\alpha^{11}-1=0$, prove that $\alpha^n$ is also a special root of this equation for all $n=1,2,..,10$. 
Considering these $11$ (distinct!) roots, conclude that the polynomial $f(x)=x^{11}-1$ factors as $f(x)=(x-1)(x-\alpha)(x-\alpha^2)\ldots(x-\alpha^{10})$. 
Substitue $x=-1$ on both sides.