How to show that $\cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n} = -1$ for all positive integers $n$? It doesn't make sense to me, how we will even show it works for $n =1$  and we can work out induction from there but how can I show even the base case.
Thanks in Advance.

If you want to prove the formula without complex analysis, here is the trick:
Let 
$$S:= \cos\frac{2\pi}{n} + \cos\frac{4\pi}{n} + \ldots+ \cos\frac{2\pi(n-1)}{n}$$
Then
$$\sin(\frac{\pi}{n}) S= \sin(\frac{\pi}{n})  \cos\frac{2\pi}{n} +\sin(\frac{\pi}{n})  \cos\frac{4\pi}{n} + \ldots+ \sin(\frac{\pi}{n}) \cos\frac{2\pi(n-1)}{n}$$
Now, as
$$\sin(\frac{\pi}{n}) \cos\frac{2 k\pi}{n} = \frac{1}{2} \left( \sin(\frac{(2k+1)\pi}{n}) -  \sin(\frac{(2k-1)\pi}{n}) \right) $$ 
you get a telescopic sum.
