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Prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using Gelfond-Schneider's theorem.

I'm interested in this problem because I knew that ${\sqrt2}^{\sqrt2}$ is a transcendental number by Gelfond-Schneider's theorem. I've tried to prove that ${\sqrt2}^{\sqrt2}$ is an irrational number without using the Gelfond-Schneider's theorem, but I'm facing difficulty. I need your help.

I crossposted to MO:

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Cannot be done. – Will Jagy Jul 31 '13 at 6:39
Meanwhile, perhaps this is what you are thinking about:… – Will Jagy Jul 31 '13 at 6:54
@Will Is there some intuition for why no irrationality proof likely exists (short of proving transcendence)? – user7530 Jul 31 '13 at 7:24
One way could be using the proof of Gel'fond-Schneider's theorem, as opposed to the Gel'fond-Schneider's theorem. – Mlazhinka Shung Gronzalez LeWy Jul 31 '13 at 7:30
@Will Jagy: It is not what I'm thinking about. I've already known its answer. I'm looking for a 'simple' proof. – mathlove Jul 31 '13 at 7:57
up vote 1 down vote accepted

I'm posting an answer just to inform that the question has received an answer by Mark Sapir on MO.

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