Q:

irrationality of $\sqrt{2}^{\sqrt{2}}$.

The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?

A:

Since this is a well-established result, this is a community wiki post.
Relevant question: Deciding whether $2^{\sqrt2}$ is irrational/transcendental
Kuzmin proved the following claim in 1930:

Theorem: If $\alpha\neq 0,1$ is algebraic, $\beta$ is positive and rational, not a perfect square, then $\alpha^{\sqrt{\beta}}$ is transcendental.

Unfortunately the paper is in Russian and I failed to find an English translation. A corollary of this is that $2^{\sqrt{2}}$ is transcendental, and so is its square root $\sqrt{2}^{\sqrt{2}}$.
The outlines of both Gelfond and Kuzmin's constructive proof can be found here. 
As David Mitra pointed out the comments, Niven's book had a section dedicated to this. I love Niven's book so much. The technique is similar to the adapted proof I posted here, proof by contradiction.
Rough idea about the construction: First assuming $\alpha^{\sqrt{\beta}}$ is algebraic. Then using sufficient large degree Lagrange interpolation polynomial to approximate $e^{(\ln \alpha)x}$ at points $\{a+ b\sqrt{2}\}$ for $a,b\in \mathbb{Z}$. Let the number of points go to infinity   the error will go to zero, this shows a transcendental function $\alpha^x$ can be interpolate using countably many algebraic points. Contradiction. 

