I'm trying to formulate Axiom of Choice in terms of $P(\prod_{i \in I}{X_i})$, but end up with the following questions.
Let $P(X)$ be the power set of $X$.In general, is it correct, or legitimate to write: $P(\prod_{i \in I}{X_i}) = \prod_{i \in I}{P(X_i)}$?
Let $|X|$ be the cardinality of $X$.Is it the case $|P(\prod_{i \in I}{X_i})| \leq \prod_{i \in I}{|P(X_i)|}$?
Note that all the elements in $\mathcal P(\prod X_i)$ are sets of functions, whereas $\prod\mathcal P(X_i)$ is a set of functions whose range is sets.
So equality never really holds.
In terms of cardinality, Ross already showed the inequality fails for finite sets. However assuming the axiom of choice we can say the following: Denote by $\mathcal P^+(X)$ all the non-empty subsets of $X$, then: $$\left|\prod_{i\in I}\mathcal P^+(X_i)\right|\leq\left|\mathcal P\left(\prod_{i\in I} X_i\right)\right|$$
This is given by the function $\langle A_i\mid i\in I\rangle\mapsto\{g\mid \forall i\in I. g(i)\in A_i\}$ is clearly injective.
The reason we need to consider $\mathcal P^+(X_i)$ is that if the empty set is present in more than one factor the above function is not an injection. Depending on the cardinality of $I$ and the $X_i$'s it might be possible to show there exists an actual injection from the product of the power sets.
Without the axiom of choice, however, things may be different. If $X_i\neq\varnothing$ but $\prod X_i=\varnothing$ then we have $\mathcal P(\prod X_i)=\{\varnothing\}$ whereas $\prod\mathcal P(X_i)$ is never empty because the function $f(i)=X_i$ is in that product (as well in the product $\prod\mathcal P^+(X_i)$).
It is even easy to show that if at least one $X_i$ has two elements then $\prod\mathcal P^+(X_i)$ has more than one element (let alone the product of the full power sets).
It might be nicer, if so, to formulate the axiom of choice as the inequality above about the product of $\mathcal P^+(X_i)$ having an injection into the power set of the product. Although to be fair I'm not sure I see much point to such formulation.