$I$ 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! 
 A: 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. 
A: 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.
A: On the basis of two answers, I present a proof in detail. The motivation for this re-formalization is that I would like to write down the proof on my own words, so that I can truly understand every tiny detail of it. All credits go to @Vera and @bang.


Lemma 1: The Cartesian product of a finite amount of countable sets is countable. (I presented a proof here)
Lemma 2: $\mathcal{P}(\mathbb N)$ is uncountable.

Let $J_1 := \{ i \in I \mid |A_i| \ge 2\}$, $J_2 :=J \setminus J_1 = \{i \in I \mid |A_i| = 1\}$, and $a_i$ be the element of $A_i$ for all $i \in J_2$. Consider
$$\begin
{array}{l|rcl}
G : & \prod\limits_{i\in J_1}A_i
 & \longrightarrow & \prod\limits_{i\in J}A_i \\
    & \phi & \longmapsto & G(\phi) \end{array}$$
where $\phi:J_1 \to \bigcup_{i\in J_1}A_i$ such that $\phi(i) \in A_i$ for all $i \in J_1$.
$G(\phi)$ is defined by
$$G(\phi)(i)=\begin{cases}\phi(i) & \text{ if } i \in J_1\\
a_i & \text{ if } i \in J_2
\end {cases}$$
a. $G$ is injective
For $\phi_1,\phi_2 \in \prod\limits_{i\in J_1}A_i$ and $G(\phi_1)=G(\phi_2)$. Then ${G(\phi_1)|}_{J_1}={G(\phi_2)|}_{J_1}$ and ${G(\phi_1)|}_{J_2}={G(\phi_2)|}_{J_2}$. Then $\phi_1(i)=\phi_2(i)$ for all $i \in J_1$ and $a_i = a_i$ for all $i \in J_2$. Thus $\phi_1 = \phi_2$. Hence $G$ is injective.
b. $G$ is surjective
For $\psi \in \prod\limits_{i\in J}A_i$, we define $\phi \in \prod\limits_{i\in J_1}A_i$ by $\phi:={\psi}_{|J_1}$. Then $G(\phi)(i)=\phi(i)$ for all $i \in J_1$ and $G(\phi)(i)=a_i$ for all $i \in J_2$. Then $G(\phi)(i)=\psi(i)$ for all $i \in J_1$ and $G(\phi)(i)=\psi(i)$ for all $i \in J_2$ (Since $A_i = \{a_i\}$ for all $i \in J_2$). Thus $G(\phi)=\psi$. Hence $G$ is surjective.
To sum up, $G$ is bijective.
Case 1: $J_1$ is finite
Then $\prod\limits_{i \in J_1}A_i$ is countable by Lemma 1. Furthermore, $G:\prod\limits_{i \in J_1}A_i \to \prod\limits_{i\in J}A_i$ is bijective. Hence $\prod\limits_{i\in J}A_i$ is countable.
Case 2: $J_1$ is infinite
Let $B_i := \{0,1\}$ for all $i \in \mathbb N$.
Since $\aleph \le |J_1|$ and $|B_i| \le |A_i|$ for all $i \in J_1$, $|\mathcal{P}(\mathbb N)|=|\prod\limits_{i \in \mathbb N}B_i| \le |\prod\limits_{i \in J_1}A_i|$. By Lemma 2, $\mathcal{P}(\mathbb N)$ is uncountable. Thus $\prod\limits_{i \in J_1}A_i$ is uncountable. Furthermore, $G:\prod\limits_{i \in J_1}A_i \to \prod\limits_{i\in J}A_i$ is bijective. Hence $\prod\limits_{i\in J}A_i$ is uncountable.
