Let $\zeta=e^{2\pi i/n}$ Prove that $x^n -1 =(x-1)(x-\zeta)(x-\zeta^2) \dots (x-\zeta^{n-1})$

Question

This is a question about cyclotomic polynomials and I have already shown that $x^n-1 =\Pi\Phi_d(x)$, taking the product over all divisors d of n.

Answer 1

Each of $\zeta^j$ for $j=0, \dots, n-1$ is a root of $X^n -1$. Thus $(X-\zeta^j)$ must divide it. Since all the $\zeta^j$ are distinct, the linear factors are co-prime and thus the product divides $X^n - 1$ too.

As the degree of both is $n$ and they are both normed, equality follows.

Answer 2

Hints:

1) Show that for $\;1\le j,\,k\le n\;,\;\;\zeta^j=\zeta^k\iff j=k\;$

2) Show that $\;\left(\zeta^k\right)^n=1\;,\;\;\forall\,k=1,2,...,n\;$

3) Show that all the roots of $\;x^n-1\;$ are different

There you go...