sequences and series - How to prove $sum_{k=1}^n cos(frac{2 pi k}{n}) = 0$ for any n>1?

I can show for any given value of n that the equation

$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$

is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction and converting the cosine to a sum of complex exponential functions

$$\frac{1}{2}\sum_{k=1}^n [\exp(\frac{i 2 \pi k}{n})+\exp(\frac{-i 2 \pi k}{n})] = 0$$

and using the conversion for finite geometric sequences

$$S_n = \sum_{k=1}^n r^k = \frac{r(1-r^n)}{(1-r)}$$

I have even tried this this suggestion I have seen on the net by pulling out a factor of $\exp(i \pi k)$ but I have still not gotten zero.

Please assist.

Answer

We have

$$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{n}\right)=\textrm{Re}\left(\sum_{k=1}^{n}e^{2\pi ik/n}\right)$$ and so $$\sum_{k=1}^{n}e^{2\pi ik/n}=\frac{e^{2\pi i/n}\left(1-e^{2\pi i}\right)}{1-e^{2\pi i/n}}$$ and notice that $$e^{2\pi i}=\cos\left(2\pi\right)+i\sin\left(2\pi\right)=1$$ so the claim follows.