TITLE: Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using a theorem
QUESTION [19 upvotes]: Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem.
We know that ${\sqrt2}^{\sqrt2}$ is a transcendental number by the Gel'fond-Schneider's theorem. I've tried to prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gel'fond-Schneider's theorem, but I'm facing difficulty. I need your help.
This question has been asked previously on math.SE without receiving any answers.

REPLY [21 votes]: You do not need to use Gelfond-Schneider theorem, you can just repeat one of the well known easy proofs of that theorem. For example, a much stronger theorem is proved in an Appendix to Lang's "Algebra", the proof is only 4 pages long. The proof uses only elementary linear algebra, some calculus and first notions of Galois theory. That proof can be significantly shortened  if you want to prove irrationality only. 
Edit  I have written a more or less complete proof here . Galois theory is not needed there (except for the fact that the norm of an integral element is an integer), but one needs the maximal modulus theorem for analytic functions. I am sure that can be avoided also. From calculus, one needs the Taylor formula (no integration is required).