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. 
 A: The only reasonable meaning of $\sqrt 2^{\sqrt2^{\sqrt2^{\cdots}}}$ would seem to be the limit of the sequence
$$ \sqrt2, \sqrt2^{\sqrt2}, \sqrt2^{\sqrt2^{\sqrt2}}, \ldots $$
which can also be defined recursively as
$$ x_0 = \sqrt 2 \\
x_{n+1} = \sqrt2^{x_n} $$
This sequence converges to the number $2$, which is rational.

Why? It is fairly easy to prove (using calculus) that if $x_n$ is less than $2$, then $x_{n+1}$ is between $x_n$ and $2$. Therefore the the sequence of $x_n$s is bounded and strictly increasing, so it must converge towards something. But by continuity the limit must be a solution to $\sqrt 2^x=x$, and the only such solutions are $x=2$ and $x=4$. The latter cannot be a limit because it would require the sequence to increase past $2$, so $2$ is the only possibility.
A: 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.
A: Assuming it converges, you can take log of both sides to get 2ln(x)=xln(2), which has solutions 2 and 4. 
