This isn't properly a calculus question,
but it's a basic issue, and it's cute.

I've discussed it elsewhere;
the best way to say it is:
``irrationals aren't closed under exponentiation''
Stated this way, it's a ``duh'',
  as $(\sqrt{2}^\sqrt{2})^\sqrt{2}=\sqrt{2}^2 = 2$

It's cute in that you don't know whether $\sqrt{2}^\sqrt{2}$ is
irrational or not,
so you don't know whether $\sqrt{2}^\sqrt{2}$
or $(\sqrt{2}^\sqrt{2})^\sqrt{2}$ is a counterexample.
...but it's not a serious result, just a cute argument
that's concealing a simply (closed under algebra) operation.

I suspect it's transcendental;
to check, take the log of everything,
and then you see that it's a question about whether
$\log p_i$ (logs of primes) are independent,
not just over $\bQ$, but over other number fields.
I suspect that they're independent over $\bar \bQ$, but have no real
basis for this.
