Showing the nth roots of unity satisfy two properties Assuming $ω_0,ω_1,...,ω_n$ are the $n^{th}$ roots of unity, I am asked to show $$(x−ω_0)(x−ω_1)···(x−ω_{n−1}) =x^n−1$$ and $$\sum^{n−1}_{a=0}ω_a= 0$$ 
I understand that by definition, the $n^{th}$ roots of unity are the roots of the polynomial $x^n-1$. I'm not sure I know the properties of the roots of unity well enough to even know where to begin.
Thank you.

Since the roots of unity are the roots of the polynomial $p(z)=z^n-1$, and since the coefficient of the $z$ term is $0$, it follows immediately that $\sum \omega_a=0$. And since the roots of unity split the polynomial completely over $\mathbb C$, it follows again at once that $p(z)=\prod (z-\omega_a)=z^n-1.$ 
