My attempt:
Suppose $A\approx A', B\approx B'$, then there exists bijections $f_A:A\to A',f_B:B \to B'$.
We need to show that $B^A\approx B'^{A'}$. That is, there is a bijection from the set $F$ of all functions from $A$ to $B$ to the set $F'$ of functions from $A'$ to $B'$.
Define function $g:F\to F'$ as: for each function from $f\in F$, for $f(a)=b$ where $a\in A$ and $b\in B$, $g(f)=f'\in F'$ send $f_A(a)\in A'$ to $f_B(b)\in B'$.
But I have trouble proving that the $g$ we defined is a bijection. I think it would be better if we describe $g$ in a more formal way instead of giving a plain sentence. So how may I prove it, or if there is some better way to do that?
Thanks.
EDIT：
Following the hint from the comment. The following is my attempt to define the inverse of $g$.
For each function $f'\in F'$. Define $h(f')$ as: for each function $f'$ in $F'$ such that $f'(a')=b'$, $h(f')$ sends $f^{-1}_A(a')$ to $f^{-1}_B(b')$. We can do this because the function $f$ is invertible.
I still think in order to prove formally that $h$ and $g$ are inverses of each other, we need to formalize the definition of both of them... So could someone tell me if we could do this? Or is there other way to verify that they are inverses?
EDIT':
Given $f\colon A\to B$, then the composition $f\circ f_A^{-1}$ is a function from $A'$ to $B$; and if we then compose $f_B$ on top of this we get a function $f_B\circ f\circ f_A^{-1}$ which is indeed from $A'$ to $B'$. Define $g: F\to F'$ as $g(f)=f_B\circ f\circ f_A^{-1}$. We prove that $g$ is a bijection.
Suppose $g(f_1)=g(f_2)$, then for every $a'\in A'$ we have that $g(f_1)(a')=g(f_2)(a')$. Thus $f_B\circ f_1\circ f^{-1}_A(a')=f_B\circ f_2\circ f^{-1}_A(a')$. Use the fact that $f_B$ is bijective, we have $f_1(f_A^{-1}(a'))=f_2(f_A^{-1}(a'))$ for all $a'\in A'$. As $f^{-1}_A$ is surjective, $f_1(a)=f_2(a)$ for each $a\in A$ and therefore $f_1=f_2$.
For each $f'\in F'$, we need to prove that there exist $f\in F$ such that $g(f)=f'$. For $f'\in F'$, define $f$ as $f^{-1}_B\circ f'\circ f_A$. Thus $f\in F$ and $g(f)=f'$ as desired. This proves $g$ is surjective.
It can be easier, and perhaps more instructive to define $g$ in a more formal way, yes.
The idea is basic. Given $f\colon A\to B$, then the composition $f\circ f_A^{-1}$ is a function from $A'$ to $B$; and if we then compose $f_B$ on top of this we get a function $f_B\circ f\circ f_A^{-1}$ which is indeed from $A'$ to $B'$.
Using the fact that $f_A$ and $f_B$ are bijections, you can prove now that also $g$ is a bijection. For example, if $g(f)=g(f')$, then for every $a'\in A'$ we have that $g(f)(a')=g(f')(a')$. Using the fact $f_A$ and $f_B$ are bijections, we can conclude that $f(f_A^{-1}(a))=f'(f_A^{-1}(a'))$ for all $a'\in A'$, and therefore $f=f'$. Similar arguments are used to show that $g$ is surjective.