Elaboration yields much conceptual insight by viewing it as a special case of a fundamental result.
The set $\,\cal O\,$ of integers $\rm\:n >0\:$ such that $\rm\:a^{\large n} \equiv 1\:$ is closed under positive subtraction, i.e.
$$\rm \color{#90f}n>\color{#0a0}m\,\in\,{\cal O}\,\Rightarrow\ \color{#c00}{n\!-\!m}\,\in\,{\cal O}\ \ \ {\rm by}\ \ \ 1\equiv \color{#90f}{a^{\large n}} \equiv a^{\large n-m}\, \color{#0a0}{a^{\large m}} \equiv a^{\large\color{#c00}{n-m}}\, $$
So, by the theorem below, every element of $\rm\,\cal O\,$ is divisible by its least element $\rm\:\ell\ \! $ := order of $\rm\,a.$
Theorem $\ \ $ If a nonempty set of positive integers $\rm\,\cal O\,$ satisfies $\rm\ n > m\, \in {\cal O} \, \Rightarrow\, n\!-\!m\, \in \cal O$
then every element of $\rm\,\cal O\,$ is a multiple of the least element $\rm\:\ell \in\cal O.$
Proof $\ $ If not there's a least nonmultiple $\rm\:n\in \cal O,\:$ contra $\rm\:n\!-\!\ell \in \cal O\:$ is a nonmultiple of $\rm\:\ell$.
This immediately yields the following very useful
Corollary $\ $ If $\,a\,$ has $\,\color{#0a0}{{\rm order}\,\ \ell}\ $ then $\,a^{\large n}\! \equiv 1\iff \ell\mid n$
Proof $\ \ (\Rightarrow)\ $ Follows by the Theorem. $\ \ (\Leftarrow)\ \ n =\ell k\,\Rightarrow\, a^{\large n}\! \equiv (\color{#0a0}{a^{\large \ell}})^{\large k}\!\equiv \color{#0a0}{\bf 1}^k\equiv 1$
Remark $\ $ See here for elaboration on the Theorem, including other proofs. For more on the key innate algebraic structure see this post on order ideals / groups and denominator ideals.