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 omegas are the roots of $x^n-1$
You can factor $x^n-1$ to get  $$x^n-1=(x−ω_0)(x−ω_1)···(x−ω_{n−1}) $$
The sum of roots of a monomial  of degree $n$ is the opposite of the coefficient of $x^{n-1}$ which in this case is $0$ 
