theorem dealing with order of numbers I'm having trouble understanding part of the proof to a theorem. The theorem states that 
Let $ord_ma = e$ and k any positive integer. Then $ord_m(a^k)$ = $e/(e,k)$.
The part I don't understand is how they conclude r|s. Here is how that part of the proof goes.
Proof: Let $ord_m(a^k) = r$ and $d = gcd(e, k)$. Then $e = sd$ and $k = td$, where s and t are positive integers such that $gcd(s, t) = 1$. Since$(a^k)^s = (a^{td})^s = (a^{sd})^t = (a^e)^t≡ 1^t ≡ 1 (mod m)$ by Theorem 10.1, $r|s$.
Theorem 10.1 just says "Let a be a positive integer such that $(a,m) = 1$ and $ord_m a = e$. Then $a^n ≡ 1 (mod m)$ if and only if $e|n$.
I do know that since $ord_ma = e$ => r|$ord_ma$ =>r|e => r|sd. But I don't get it from here how they get r|s?  

The expression is $(a^k)^s\equiv 1$.  Set $b=a^k$, so you have $b^s\equiv 1$.  By Theorem 10.1, the order of $b$ divides $s$.  But the order of $b$ is $r$.  So $r|s$.
