The problem is at the very first step.
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Let ... No let's not.
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This is actually false. For x = √2, x^(x^...) is actually 2. You can see for yourself: evaluate √2^√2 and then type √2^ANS and keep pressing the equal sign to see where it goes...
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Let 2=4. So, 4=2. QED. In all serious I assume there is some kind of divergence for this, but I have no education on infinite exponents. I don't think there could be a number y that could converge to anything other than 0 or 1, but then again, I'm acting only on my intuition.
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There is a similar proof about 1 being the largest integer, that only works if you start with the assumption that "largest integer" is something that can exist.
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I think this falls into the category of no solution existing for the equation x^x^x... = 4. It's like doing row reduction on a matrix where the final row is all zeros where the augmented side is say, 4. Like 0 0 0 | 4. And then instead of claiming no solution you claim, 0 = 4.
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It's because z^z^z^z^... doesn't converge to any number greater than e, which is almost 2.718281828(and z can just be in interval [0,(e)^(1/e)]), so there exists x^x^x^x... = 2, but no y^y^y^y^... = 4 .
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the issue is that while x^x^x... = 2 has an actual solution of √2, the function can't equal 4 cuz the function has a maximum convergent result of e for x = e^(1/e), so the presumption made is false so the proof is invalid. It's like going "let x+2=x+4" which is obviously wrong
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