How to show that $\bar{a}^t \equiv (\bar{a}^ta^t)(a^t)$? The idea is to show that $ord_m a= ord_m \bar{a}$. It is in the following sequence of modulo's?
$ord_m a$ exists => $a^t \equiv 1 \pmod n$. Then $\bar{a}^t \equiv (\bar{a}^ta^t)(a^t) \equiv (\bar{a}a)^t a^t \equiv 1^t \cdot 1 \equiv 1 \pmod n$.
Another possibility is that it is an error there. You can check whole proof from http://www.aw-bc.com/rosen/Rosen_NumTheory5_SSM.pdf -> 9.1.9. at page 79. Because $\bar{a}$ is the inverse of a modulo m <=> $a\bar{a} \equiv 1 \pmod n$.