$I$ 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! 

The Cartesian product will be countable if any of the sets $A_i$ is empty, since then the Cartesian product will also be empty. Let us assume that for all $i$, $A_i\neq \varnothing$. 
If there exists a finite subset $J$ of $I$ such that $A_i$ is a singleton for all $i\in I\setminus J$, then the Cartesian product will be countable. To see this, we can rearrange the terms to assume that $A_1, \ldots, A_n$ have more than one element, but $A_{n+1}, A_{n+2}, \ldots$ each have exactly one element. Then there is a bijection between $\prod_{i=1}^\infty A_i$ and $\prod_{i=1}^n A_i$. 
Now let us consider the case where infinitely many of the $A_i$ have at least two elements (still assuming each $A_i$ is non-empty).  Then $\prod_{i=1}^\infty A_i$ will contain a subset which has the same cardinality at $\{0,1\}^\mathbb{N}$. We should be able to answer the question from here. I will leave the details to you. 
