Let the order of $a$ modulo $n$ be $k$. Show that inverse of $a$ modulo $n$ is
$$ a^{k−1} \pmod n $$
That $k$ is the order of $a$ modulo $n$ means
$$ a^k \equiv 1 \pmod {n} $$
Factorising
$$ a \times a^{k-1} \equiv 1 \pmod {n} $$
By definition of inverse, this tells us the inverse of $a$ modulo $n$ is $a^{k-1}$.