Let $A$ be a countable set.
Use the axiom of choice to prove the existence of a function sequence $\langle h_{k} \space | \space 1 \leq k \in \mathbb{N} \rangle$ s.t for each $k$,
$h_k: \mathbb{N} \rightarrow A^{k}$ is a bijection.
Well, I know that since the set $A$ is countable then if we consider the set of all bijections from $\mathbb{N}$ to $A^{k}$ for some $k$, (which exists by separation), it isn't empty and we can pick one using the axiom. Specifically, I struggle with the technique part, which is to figure what should be the domain set for the choice function. I only could think of taking $\mathbb{N}$ as the domain set and this way we can create a sequence but I cant see how to it link with A elements.
I'd like to have some advice. (I know it is also can be proved using recursion but I'm asked to use the axiom of choice particularly).
Thanks
You don't actually need the axiom of choice here, of course. You can simply construct these bijections by composing and iterating the bijection between $\Bbb N$ and $A$, as well as $\Bbb N$ and $\Bbb N^2$.
The exercise, I suppose, is to argue that $A^k$ is countable for all $k\in\Bbb N$, and it is countably infinite when $k>0$. Which means that the set $B_k=\{f\colon\Bbb N\to A^k\mid f\text{ is a bijection}\}$ is non-empty. Therefore, $\{B_k\mid 1\leq k\}$ is a family of non-empty sets, so using the axiom of choice, we can choose $h_k\in B_k$, which is exactly to say that there is a sequence $\langle h_k\mid 1\leq k\rangle$ as wanted.
Of course, you seem to have noticed that, and you also seem to have noticed that using the axiom of choice here is unnecessary.