Let $\omega_n=e^{2\pi i/n}$. Prove $1+\omega_n+\omega^2_n+\cdots+\omega^{n-1}_n=0.$ 
Let $\omega_n=e^{2\pi i/n}$. Prove $1+\omega_n+\omega^2_n+\cdots+\omega^{n-1}_n=0.$

Note: I tried using mathematical induction and some basic trig identities but I can't seem to crack this one. Any hints???
 A: HINT: Notice that $\omega_n^n=1$ so $\omega_n$ is a root of $x^n-1$. Also,
$$x^n-1=(x-1)(1+x+x^2+...+x^{n-1})$$
Thus, since $\omega_n-1\ne 0$, you can conclude that...
A: Note that
$\omega_n = e^{2\pi i / n} \tag 1$
satisfies
$\omega_n^n - 1 = (e^{2\pi i / n})^n - 1$
$= e^{2 \pi n i / n} - 1 = e^{2 \pi i} - 1 = 1 - 1 = 0; \tag 2$
we also have the well-known identity
$\omega_n^n - 1 = (\omega_n - 1)(1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1})$
$= (\omega_n - 1) \displaystyle \sum_0^{n - 1} \omega_n^k; \tag 3$
combining (2) and (3) yields
$(\omega_n - 1)(1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1})$
$= (\omega_n - 1) \displaystyle \sum_0^{n - 1} \omega_n^k = 0, \tag 4$
and since 
$\omega_n \ne 1, \tag 5$
we have
$\omega_n - 1 \ne 0; \tag 6$
now dividing (4) through by $\omega_n - 1$ yields
$1 + \omega_n + \omega_n^2 + \ldots + \omega_n^{n - 1} =  \displaystyle \sum_0^{n - 1} \omega_n^k = 0, \tag 7$
the desired result.
A: $$1+\omega_n+\omega_n^2+\cdots+\omega_n^{n-1} = \frac{1-\omega_n^n}{1-\omega_n}=\frac{1-e^{2\pi i}}{1-\omega_n}=\frac{1-1}{1-\omega_n}=0.$$
