Show that if $r$ is an nth root of $1$ and $r\ne1$, then $1 + r + r^2 + ... + r^{n-1} = 0$.

Multiply top and bottom by $(1-r)$.

$$1+r+r^2+\cdots+r^{n-1}=\frac{(1+r+r^2+\cdots+r^{n-1})(1-r)}{(1-r)}=\frac{1-r^n}{1-r}=\frac{0}{1-r}=0$$

Note: $r\neq 1$ is necessary, else the statement is false.

$1 + r + r^2 + r^3 + ... + r^{n-1}$ is the sum of the first $n$ terms of a geometric series. The formula for this summation, as long as $r \ne 1$, is

$$\frac{r^n - 1}{r - 1}$$

Since $r$ is an nth root of unity, $r^n - 1 = 0$. Since the denominator is nonzero, this quantity is equal to zero.

Hope this helps!

We know that $r^n - 1 = 0$ but we can factorise the LHS as $(r-1)(r^{n-1}+r^{n-2}+...+r+1)=0$.

However $r\neq 1$ so the other bracket must be $0$.