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.

The right hand side should be

$$\frac{1-e^{\frac{in2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}}=0$$

since $e^{\frac{in2\pi}{n}}=e^{i2\pi}=1$.

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.

Since $c \neq 1$ we have $1+c+\cdots+c^{n-1} = { 1-c^n \over 1-c}$. Then $c^n=1$ finishes.

Are you sure you are supposed to use complex analysis to do this? If so what I am saying is irrelevant but here is another 'easy' method.

If $c$ is an nth root of unity then $c$ satisfies the equation $c^n-1$. Factorise this and you should be able to get to the answer.

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., $$(e^{2\pi i/n}-1)\sum^{n-1}_{k=0} e^{2k\pi i/n} = 0.$$