Induction suffices to show that (oo)^n is countable for every
n a natural number, but that does not suffice to show that (oo)^(oo) is countable, because the power is not a natural number. The induction you have only proves the step that if X^n is countable then X^(n+1) is countable. That never proves the case X^(oo)
There is a thing called transfinite induction, and there you have to show the separate case that if predicate P holds for all n_i then it holds for the limit. As stated elsewhere you can't do that in this case because it's not true. 2^(oo) is uncountable.
> Am I missing something here?
Yes. You are forming perfectly reasonable conjectures based on limited experience, but those conjectures turn out to be false when you study the subject in its own right, instead of just those bits you need for the limited version of calculus that you've done. It's a problem I often have when teaching people this stuff that they have only ever dealt with "nice" functions and "easy" situations. I wrote a little about that here:
http://www.solipsys.co.uk/new/PokingTheDustyCorners.html?HN_...