Sum of nth roots of unity Question: If $c\neq 1$ is an $n^{th}$ root of unity then, $1+c+...+c^{n-1} = 0$
Attempt: So I have established that I need to show that $$\sum^{n-1}_{k=0} e^{\frac{i2k\pi}{n}}=\frac{1-e^{\frac{ik2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}} =0$$ by use of the sum of geometric series'. My issue is proceeding further to show that this is indeed true.

If a finite set of complex numbers is symmetric about a line passing through the origin, then its sum must lie on that line; if it is symmetric about two different lines through the origin, then its sum must be zero. The $n^{\text{th}}$ roots of unity are the vertices of a regular $n$-gon centered at the origin, which has $n$ lines of symmetry. Hence, for $n\ge2,$ the sum is zero.
