While I haven't found a proof that sqrt(2)^(sqrt(2)^sqrt(2))√2√2√2 is irrational, here are answers to the first and third parts.
sqrt(2)^sqrt(2)√2√2 is irrational:
The Gelfond-Schneider theorem states that given algebraic numbers a, ba,b where a != 0, 1a≠0,1 and bb is irrational, a^bab is transcendental.
As sqrt(2)√2 is a root of x^2-2x2−2, it is algebraic. It is well known that sqrt(2)√2 is irrational, although this fact may be proven by supposing that sqrt(2)=p/q√2=pq with p, q in ZZ, squaring both sides, multiplying by q^2, and then noting that the left hand side will have an odd exponent for 2 in its prime factorization, while the right hand side will have an even exponent, a contradiction.
By the above, sqrt(2)^sqrt(2) fulfills the conditions for the theorem, and thus is transcendental (and therefore irrational).
sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) = ""^oosqrt(2) is rational:
We will prove the stronger result that ""^oosqrt(2)=2.
First, to show that ""^oosqrt(2) converges, we will show that the sequence a_n = ""^nsqrt(2) is monotone increasing and bounded above.
We proceed by induction. As our base case, note that
a_1 = sqrt(2) < sqrt(2)^sqrt(2) = a_2 < sqrt(2)^2 = 2. Now, suppose that for some positive integer k, the sequence a_1, a_2, ..., a_k is increasing and a_k < 2. Then
a_k = sqrt(2)^(a_(k-1)) < sqrt(2)^(a_k) = a_(k+1) < sqrt(2)^2 = 2
Thus, by induction, a_n is increasing and bounded above by 2.
Now that we know ""^oosqrt(2) converges, let a= ""^oosqrt(2). Then sqrt(2)^a = a, meaning a is a root of f(x) = sqrt(2)^x - x. Finding the critical points of f(x), we have
f'(x) = 1/2ln(2)sqrt(2)^x - 1
=> f'(x) = 0 <=> x = log_sqrt(2)(2/ln(2))
As f(x) has only one critical point, it can have at most two roots. By observation, 2 and 4 are both roots of f(x), meaning a in {2, 4}. But, by the above work, a<=2, meaning a=2.
