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.

Yet another answer, with more general applicablity:
For any polynomial $p(x)=a_n x^n + a_{n-1}x^{n-1}+\cdots+a_1 x +a_0$, with roots $x_1,x_2,\ldots,x_n$, the sum of the roots is given by $x_1+x_2+\ldots+x_n=-\frac{a_{n-1}}{a_n}$.
This is because $p(x)=a_n(x-x_1)(x-x_2)\cdots(x-x_n)$, and you can work out from there that the coefficient of $x^{n-1}$ is minus the sum of the roots, times the $a_n$ prefactor. (Notice it's also easy to get the product of the roots, as well as the sum of products of any fixed number of terms from $1$ to $n$).
In your case you're looking at the root of the polynomial $p(x)=x^n-1$, so $a_n=1,a_0=-1$ and $a_k=0$ for other $k$. In particular, unless $k=1$, you have $a_{n-1}=0$ so the sum of the roots is zero.
