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

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...

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.

$$1+\omega_n+\omega_n^2+\cdots+\omega_n^{n-1} = \frac{1-\omega_n^n}{1-\omega_n}=\frac{1-e^{2\pi i}}{1-\omega_n}=\frac{1-1}{1-\omega_n}=0.$$