Is $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}$ rational, algebraic irrational, or transcendental? Let 
$$
x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}.
$$
Assume $x$ is algebraic irrational. By the Gelfond-Schneider Theorem, $x^x$, which is also $x$, is transcendental, a contradiction.  
But I have no idea how to do the rest. 

Your number is the limit of the sequence $a_0=1$ and $a_{n+1}=(\sqrt 2)^{a_n}=2^{a_n/2}$
This sequence is increasing and bounded.
So its limit is the solution of the equation $a=2^{a/2}$
From where you get $a=2$ which is rational.
