User blog comment:Edwin Shade/A Proof/@comment-1605058-20171104122735/@comment-1605058-20171109220024

Since I don't want to write up all the details of the proof of power tower convergence in whatever interval, let me just give you the ideas of what goes into the proof.

The most important tool here is the - it gives us a sufficient criterion for an infinite sequence of reals to converge. The easier case is when \(1\alpha\) and vice versa. However, we can verify that if \(x>\alpha\), then \(x>a^{a^x}>\alpha\). Therefore the subsequence \(a_{2n}\) is decreasing and bounded, hence it converges to some number \(\beta\), which satisfies \(a^{a^\beta}=\beta\). Since \(a^{a^\alpha}=a^\alpha=\alpha\), we can only conclude that \(\beta=\alpha\) if we knew that \(a^{a^x}=x\) has exactly one solution, and this is in fact the case for \(a\geq e^{-e}\). Hence \(a_{2n}\) converges to \(\alpha\), hence \(a_{2n+1}=a^{2n}\) converges to \(a^\alpha=\alpha\). We can now finally conclude the whole sequence \(a_n\) converges to \(\alpha\).