Sum of n-th roots of unity

Question

I'm being asked to prove that
$$1 + \omega + \omega^2 + ... + \omega^{n-1} = 0$$ where $\omega \ne 1$ is an n-th root of unity, and I don't know where to start

I feel like there's something terribly obvious that I'm missing here.

Answer 1

$$x^n-1=0$$ $$(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)=0$$ Then, $x=1$ or $x^{n-1}+x^{n-2}+\cdots+x+1=0$.

The root of the second equation is what we denote by $\omega$.

Answer 2

Hint : multiply by $\omega - 1 $

Answer 3

Multiply both sides by $(1-w)$, then you will come up with an expression $1-w^n$, which must be zero.