If the sets $A_i$ are all countably infinite, then the product $A$ is also countably infinite. The argument is like this: if the product $A$ is countably infinite, then you can enumerate all of the elements of $A$ as a countable sequence like this one:
$$x_1=\langle a_1, b_1,c_1,d_1,\ldots\rangle\\ x_2=\langle a_2, b_1,c_1,d_1,\ldots\rangle\\
(The exact order doesn't matter; all that matters is that all the elements in $A$ appear in this sequence, by asumption.)
You can use a diagonal argument to define a member of $A$ that doesn't appear in this list. Because the list is supposed to contain all of the elements of $A$, this will be a contradiction; therefore $A$ is uncountable.
Define the member $y=\langle a,b,c,d,\ldots\rangle \in A$ as follows. For the first item in $y$, pick some element of $A_1$ that's different from the first item in $x_1$. This ensures that $y\neq x_1$. For the second item in $y$, pick some element of $A_2$ that's different from the second item in $x_2$. This ensures that $y\neq x_2$. And so on.
The resulting member $y$ is a member of the product $A=A_1\times\ldots$ because we chose its first item from $A_1$, its second item from $A_2$, and so on. However, $y$ is different from every element in the countable list of $x_i$. (Each $x_i$ has a different $i$th item .) Therefore $y$ is a member of $A$ that was missed by the countable enumeration. Therefore $A$ is uncountable.
You can extend this argument to the case where most, but not all, of the $A_i$
are infinite. $A$
will be uncountable whenever infinitely many $A_i$
are infinite. Otherwise, $A$
will be countably infinite.