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.

A variation of the answer of bof: the $n$-th roots of unity are the vertices of a regular $n$-gon centered at the origin and the product by $e^{2\pi i/n}$ is a rotation of angle $2\pi/n$ that leaves invariant the set of $n$-th roots. This means that
$$e^{2\pi i/n}\sum^{n-1}_{k=0} e^{2k\pi i/n} = \sum^{n-1}_{k=0} e^{2k\pi i/n}$$
i.e.,
$$(e^{2\pi i/n}-1)\sum^{n-1}_{k=0} e^{2k\pi i/n} = 0.$$
