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. 

Assuming it converges, you can take log of both sides to get 2ln(x)=xln(2), which has solutions 2 and 4. 
