If $\omega=\cos(2\pi/n)+i\sin(2\pi/n)$, what are the distinct $n$th roots of 1?

Question

My work so far:

$\omega^1=\cos(2\pi/n)+i\sin(2\pi/n)=e^{i(2\pi/n)}$

$\omega^2 =e^{i(4\pi/n)}$

$\omega^{n-1} =e^{i(2(n-1)\pi/n)}$

$\omega^n =e^{i(2\pi)}$

These appear to be the distinct $n$th roots, but I am having troubles determining how to explicitly show that they are.

Answer 1

For $0\le k\lt n$, $\left(e^{2\pi ik/n}\right)^n=e^{2\pi ik}=1$, so they are all $n^\text{th}$ roots of $1$.

Given $0\le j\lt k\lt n$ $$ \frac{e^{2\pi ik/n}}{e^{2\pi ij/n}}=e^{2\pi i(k-j)/n}\ne1 $$ so they are all different.

Answer 2

$(\omega^k)^n=(\omega^n)^k=1^k=1$