Showing $\operatorname{ord}_ma^s=k$ if $\gcd(s,k) = 1$ and $\operatorname{ord}_ma=k$

If $\operatorname{ord}_ma=k$ and if $\gcd(s,k)=1$ for some $s\ge1$, prove that $\operatorname{ord}_ma^s=k$.

I know that $a^n\equiv 1 \pmod m$ for $n \ge 1$ if and only if $k\mid n$.

However, I don't know how to use this.

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If the order of $a^s$ is $r$, then $k$ divides $rs$, but $\gcd(k,s)=1$, so $k$ divides $r$. On the other hand, it's clear that $r$ is at most $k$.