Direct sum countably infinite

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Let $\{A_i:i\in N\}$ be a collection of nonempty sets. The direct sum of the collection is $\bigoplus_{i\in N}A_i=\bigg\{(a_1,a_2,a_3,\ldots)\in\prod_{i\in N}A_i:a_i\neq0 \text{ for only finitely many }i\bigg\}.$

Let $\{A_i:i\in N\}$ be a collection of countably infinite sets. Show that $\bigoplus_{i\in N}A_i$ is countably infinite.

Attempt at solution: I realize that if the number of non-zero coordinates is finite, then one can order the coordinates in such a way that the natural numbers $N$ map onto the direct sum. I am new to the notation, though, and am having trouble constructing a formal proof. Would this involve some function from $N\times N\times N\times...$?

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Hint: First show that for any $n \in N$, the subset $A^{(n)} = \{ (a_i) \in \bigoplus_{i \in N} A_i : a_k = 0 \text { for } k > n\}$ is countable, and then note that $\bigoplus_{i \in N} A_i$ is precisely the ascending countable union of countable sets $\bigcup_{n \in N} A^{(n)}$.