$I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i \neq k$. Is $\prod\limits_{i\in I}A_i$ countable? The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i \text{ for all } i \in I\}$$

Let $(A_i \mid i \in I)$ be a family of non-empty indexed sets where $I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i \neq k$. Is $\prod\limits_{i\in I}A_i$ countable?

I found that it's not too hard to conclude when $I$ is finite or when $A_k$ is uncountable. Please give me some hints in this case! 

If $J:=\{i\in I\mid A_i\text{ contains at least two elements }\}$ then the product contains a subset that has the same cardinality as $\{0,1\}^J$ which on its turn has the same cardinality as powerset $\wp(J)$. 
This set is not countable if $J$ is infinite, so in that case the product is not countable.
If $J$ is finite and for every $i\in J$ the set $A_i$ is countable then the product is countable.
If for at least one $j\in J$ the set $A_i$ is not countable then the product is not countable.
If $J$ is empty then the product is a singleton.
