How to prove $|a^k|=n/\gcd(n,k)$ whenever $|a|=n$?

You have two things to show. Namely that:

1. $(a^k)^{n/\gcd(n,k)}=e$ (where $e$ denotes the identity)

and that $n/\gcd(n,k)$ is the smallest positive power $p$ of $a^k$ such that $(a^k)^p=e$:

2. For all $m>0$: $(a^k)^m=e \implies n/\gcd(n,k)\leq m$

The first part is easy:

$$ (a^k)^{n/\gcd(n,k)}=(a^n)^{k/\gcd(n,k)}=e^{k/\gcd(n,k)}=e $$

For the second part, let $m\in\mathbb{N}$ be such that $(a^k)^m=a^{km}=e$. Since the order of $a$ is $n$, it follows that $n\mid km$. Therefore we also have

$$ \frac{n}{\gcd(n,k)}\mid \frac{k}{\gcd(n,k)}m $$

Now $\gcd(n/\gcd(n,k),k/\gcd(n,k))=1$ (try to prove this), so it follows that

$$ n/\gcd(n,k)\mid m $$

Hence in particular $n/\gcd(n,k)\leq m$.

The order of $a^k$ is the smallest $r>0$ such that $a^{kr}=e$, i.e. such that $kr$ is a multiple of the order $n$ of $a$. This means $kr$ is the least common multiple of $k$ and $n$.

As $\operatorname{lcm}(k,n)=\dfrac{kn}{\gcd(k,n)}$, we have: $\quad r=\dfrac{n}{\gcd(k,n)}.$

The proof is simpler (and more general) if using basic gcd / lcm laws vs. Bezout gcd identity.

Lemma $\ \ (a^{\large k})^{\large j}\! = 1\color{#90f}{\underset{a^n\,=\,1}\Longleftarrow}\!\!\!\!\!\color{#0a0}{\overset{{\rm ord}\,a\,=\,n}{\Longrightarrow}} n/(n,k)\mid j\,.\ \ $ Proof:

$$ (a^{\large k})^{\large j}\! = 1\color{#90f}{\underset{a^n\,=\,1}\Longleftarrow}\!\!\!\!\!\color{#0a0}{\overset{{\rm ord}\,a\,=\,n}{\Longrightarrow}} n\mid kj\!\iff\! n\mid nj,kj\!\color{#c00}\iff\! n\mid(nj,kj)\!=\!(n,k)j\!\iff\! n/(n,k)\mid j\qquad$$

$\,\color{#0a0}{\rm First\ (\Rightarrow)}\,$ is by $\,n = {\rm\ ord}\, a,\,$ $\rm\color{#c00}{third}$ by the definition/universal property of the gcd and the gcd distributive law. $\,\color{#90f}{\rm Note}$ that the proof of the $\color{#90f}{(\Leftarrow)}$ chain needs only that $\,\color{#90f}{a^n =1},\,$ not $\,\color{#0a0}{{\rm ord}\,a = n},\,$ i.e. $\,a^n = 1\,\Rightarrow\, (a^k)^{n/(n,k)} = 1,\,$ or use a direct proof by writing the expt as $\,n\:\!(k/(n,k))\:\!$.

Alternatively we can use lcm instead of gcd, using notation $\,[x,y] :={\rm lcm}(x,y),$

$$ (a^{\large k})^{\large j} = 1\iff n\mid kj\iff n,k\mid kj\iff [n,k]\mid kj\iff [n,k]/k\mid j\qquad$$

Both proofs are equivalent by $\ [n,k]/k = n/(n,k),\ $ i.e. $\ [n,k](n,k) = nk\ $ [gcd $*$ lcm law]

Corollary $\ \ \bbox[5px,border:1px solid #c00]{{\rm ord}(a) = n\,\Rightarrow\,{\rm ord}(a^{\large k}) = \dfrac{n}{(n,k)} = \dfrac{[n,k]}k}\ $ since, generally

$\qquad\qquad\quad\underbrace{ b^{\large j} = 1\iff i\mid j}_{\small \textstyle \text{Lemma has }\, b = a^k}\ $ is equivalent to $\,b\,$ has order $\,i.\,$