Is it true that $\forall n \in \mathbb Z^+ \land \forall i \in \mathbb Z \land i \geqslant 0: n^i \in \mathbb Z^+$?
The reason I ask is because this would allow you to create a countably infinite set $$\forall k \in \mathbb Z \land k \geqslant 0, \{a_k \in \mathbb Z^+ : a_k = 10^k\}$$
This set would contain elements of the form $\{1, 10, 100, 1000, \ldots\}$, i.e. each element $a_k$ of the set would be a $1$ followed by $k$ zeros, up to an infinite amount of zeros. But a natural number cannot have an infinite number of digits, so that would seem to say that the natural numbers are not closed under exponentiation. Am I getting something wrong here?
They are closed under exponentiation, the contradiction you gave isn't a real contradiction.
When you say that there can be "up to an infinite amount" of zeroes, that just means that there is no upper bound on the number of trailing zeroes. However any specific number you choose from that set will have some finite number of trailing zeroes.
It's similar to how the natural numbers "go to infinity", even though every natural number you choose will have to be finite.