trigonometry - Prove that $sumlimits_{n=1}^N cos(2pi n/N)= 0$?

Is there a way to algebraically prove that $\sum\limits_{n=1}^N \cos(2 \pi n/N) = 0$ for any $N > 0$? (And if so, how?)

Answer

Hint :

$$\sum_{n=1}^N\cos\left(\frac{2\pi n}{N}\right)=\Re\left(\sum_{n=1}^N\exp\left(\frac{2i\pi n}{N}\right)\right)$$

and also :

$$\forall\theta\in\mathbb{R},\,\forall n\in\mathbb{N},\,\exp(i n\theta)=\left[\exp(i\theta)\right]^n$$