Let $\displaystyle \zeta = e^{2\pi i/n}$

Prove that

$\displaystyle x^n-1=(x-1)(x-\zeta)(x-\zeta^2)...(x-\zeta^{n-1})$

And if n is odd, that

$\displaystyle x^n+1=(x+1)(x+\zeta)(x+\zeta^2)...(x+\zeta^{n-1})$

If i could get help on the first part i think i should be able to do the odd part.