> is an infinity to an infinitieth power still countable? It shouldn't be. You can use the same principle as showing that reals aren't countable: Suppose you

> is an infinity to an infinitieth power still countable? It shouldn't be. You can use the same principle as showing that reals aren't countable: Suppose you have an ordered (infinite) set of the infinite-dimensional vectors. Doesn't matter the ordering. Then construct another vector by the following: the Nth dimension of the vector is different from the Nth dimension of the Nth element of the set. Could be +1, could be -1, could be whatever, long as it's different. Note that this doesn't actually require infinity^infinity, it's just {any set of >1 element}^infinity. The new vector is, by definition, different from every element of the set, as the Nth dimension's value is different than the Nth element of the set's Nth value, for all N. Hence the new vector isn't in the set. (Alternatively, you can biject {0-9}^infinity with the reals between 0 and 1. And the reals between 0 and 1 aren't countable.)