Let $w= \cos\frac{2k\pi}{n} + i \sin\frac{2k\pi}{n}$ be a primitive $n^{\text{th}}$ root of unity, ie, $w^n=1$ and $w^m \neq 1$ for $m \leq n$. Then the powers $$1, w, w^2, \ldots, w^{n-1}$$ are all the $n^{\text{th}}$ roots of unity.
Comments: I tried by use the fact that if $w_0 = (\cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n})$ then the $n^{\text{th}}$ roots of unity are $1, w_0, w_0^2, \ldots, w_0^{n-1}$. This case is the case in that $k=1$ in the problem. The way forward would be this using $\mod{n}$? Or there would be one where I would not need the result I commented.
Here is roadmap:
$1, w, w^2, \ldots, w^{n-1}$ are $n$th roots of unity.
$1, w, w^2, \ldots, w^{n-1}$ are all different. This is the crucial point.
There are exactly $n$ complex $n$th roots of unity.