Proof that $t^m=t^{j}$ if $t$ is an $r^{th}$ root of unity such that $r \mid k$.

Question

I need help with the following proof.

Let $j$ = $0,1,\ldots, k-1$. Also, let $t$ be an $r$th root of unity other than $t=1$ such that $r \mid k$. We know $m=j\pmod k$.

Furthermore, $m$, $j$ and $k$ are positive integers.

Prove that $t^m=t^j$ for all $j$.

It seems like this result makes sense but the use of $r \mid k$ in the proof confuses me.

Answer 1

Because $r \mid k$, let $k=lr$, then $t^k=t^{lr}=(t^r)^l=1$

Because $m=j(\text{mod }k)$, we can let $m=nk+j$, then $t^m=t^{nk+j}=t^{nk}t^j=(t^k)^nt^j=1\cdot t^j=t^j$.