User blog comment:Edwin Shade/In The Pursuit Of Organization/@comment-1605058-20171030102151/@comment-32876686-20171030155002

It does, because $$a^{a^{a^{.^{.^{.}}}}}$$ I set equal to x, therefore $$a^x$$ is still equal to an infinite power tower of a's. As there are an infinite number of a's it doesn't matter that we added one more a to this power tower because $$\infty+1=\infty$$. (Here I using infinity as a cardinal, not an ordinal.)

An example of how this works can be shown be trying to solve the equation $$x^{x^{x^{.^{.^{.}}}}}=2$$.

To solve this we just take the fact that $$x^{x^{x^{.^{.^{.}}}}}=2$$ and then say if $${x^2}=2$$ has a solution then there exists a solution for $$x^{x^{x^{.^{.^{.}}}}}=2$$. There clearly is a solution for $${x^2}=2$$, which is $$+\sqrt{2}$$, which if you evalute an infinite power tower of $$+\sqrt{2}$$'s from the top down you will find it does converge at 2.

The proof that $$a\uparrow\uparrow\infty$$ is finite for $$a\leq{e^{\frac{1}{e}}$$ is there, more or less, but could be phrased better, which I'll work on.