We know the Cartesian product of a finite number of $\mathbb{Z}^+$ is countable, for example, $\mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+$, because, let $m, n, p$ be from each of the $\mathbb{Z}^+$, we can find a one-to-one function $2^m 3^n 5^p$ that maps them to a subset of $\mathbb{Z}^+$. But what if we increase the number of $\mathbb{Z}^+$ to (countable) infinity, i.e.
$$ \mathbb{Z}^+ \times \mathbb{Z}^+ \times \mathbb{Z}^+ \times \cdots $$ ?
For your infinite product, map $(n_1,n_2,\ldots,)$ to the subset $\{n_1,n_2,\ldots\}$. Then you will get a surjection onto ${\mathbb P}({\mathbb Z}^+)$, which has cardinality greater than the cardinality of ${\mathbb Z}^+$ by Cantor's theorem. Thus your infinite product has cardinality greater than or equal to that of ${\mathbb P}({\mathbb Z}^+)$, and it is in fact equal, which can be shown with a little more work. It is the same cardinality as the set of real numbers, which can also be shown with a little more work. What is true is that the union of countably many countable sets is countable, but not the product.
