Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the [Gelfond–Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_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.


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Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using a theorem
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](https://marksapir.wordpress.com/2013/08/01/short-proof-that-2sqrt2-is-not-rational/). 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).


