Proving if $b^k = a$ and $\text{ord}(a) = n$ then $\text{ord}(b) = kn$.

Let $(G,e)$ be a group and $a \in G$ have finite order

$\quad \text{ord}(a) = n$

and let $\langle a \rangle$ denote the cyclic group generated by $a$.

Suppose for $b \in G$ and $k \ge 2$

$\quad b,\dots, b^{k-1} \notin \langle a \rangle$ and $b^k = a$

Then the order of $b$ is $kn$.

Proof

The order of $b$ must be a multiple of $n$ since $\langle a \rangle \subset \langle b \rangle$.

The order of $b$ must divide $kn$ since $b^{kn} = e$.

All that remains is to identify $kn$ distinct elements in $\langle b \rangle$.

Consider the mapping

$\quad (u,v) \mapsto a^u b^v \quad \text{where } 0 \le u \lt n \land 0 \le v \lt k$

Our work will be complete if we can show that this mapping is injective. This is accomplished by using the fact that the $b^v$ can never be a non-trivial inverses for any elements in $\langle a \rangle$.

Suppose $a^u b^v = a^s b^t$ and $u = s$. Then $v$ must be equal to $t$.

So assume, without loss of generality, that $u \gt s$. Then we can write

$\quad a^w b^v = b^t$

with $0 \lt w \lt n$.

If $v = t$ we have a contradiction since $a$ has order $n$.
If $v \gt t$ we have a contradiction since we can't construct a non-trivial inverse.
If $v \lt t$ we have a contradiction since $b^{t-u} \notin \langle a \rangle$.

This completes the proof.

Is this a valid proof?

It seems fine to me but the reason for posting this question is I could not find this on the internet of math facts. I could not find this (fact?) as a duplicate question on this site or anywhere else.

So any links to the literature that uses this would be of interest.

Your proof is correct. A far shorter proof follows.

For $1\le x<kn$, $b^x\ne e$ since $$b^x=(b^k)^{\lfloor x/k\rfloor}b^{x\bmod k}=a^{\lfloor x/k\rfloor}b^{x\bmod k}$$ Either $x\bmod k>0$ and $b^x\not\in\langle a\rangle$ because the first factor is in $\langle a\rangle$ and the second is not, or $x\bmod k=0$ but $\lfloor x/k\rfloor\in[1,n)$ and $b^x=a^{\lfloor x/k\rfloor}\ne e$.

Yet it is easy to show that $b^{kn}=e$. Hence the order of $b$ is $kn$.