Why $\mathbb Z_+^\omega\neq\cup_{n=1}^\infty\mathbb Z_+^n$? On the one hand we know that an infinite product of countable sets is uncountable, so $\mathbb Z_+\times\mathbb Z_+\times\ldots=\mathbb Z_+^\omega$ is uncountable. 
On the other hand, the finite product of countable sets is countable and the countable union of countable sets is countable, so $\displaystyle\cup_{n=1}^{\infty}\mathbb Z_+^n$, where $\mathbb Z_+^n=\mathbb Z_+\times\ldots\times\mathbb Z_+$ ($n-$times), is countable.
So this is my question: why $\mathbb Z_+^\omega\neq\cup_{n=1}^\infty\mathbb Z_+^n$?
It seems to me that, since the union's index runs to infinity, these sets should be the same.

There not even of the same type: $\cup_{n=1}^\infty \mathbb{Z}^n_+$ is a set of vectors that all have finite length: each $x$ in the union must be in some $\mathbb{Z}^n_+$ and is thus of the form $(a_1,\ldots, a_n)$ for that $n$. A vector has a unique length so it is in exactly one such set. While an $x \in \mathbb{Z}_+^\infty$ is by definition a sequence (countably infinite vector) of the form $(a_1, a_2 ,a_3, \ldots)$. So a member of the right hand side is never in the left hand side or vice versa. So certainly no equality of sets (where $A = B$ for sets $A,B$ by definition means that $x \in A$ iff $x \in B$).
