Is $e^{e^9}$ an integer? I mean, of course $e^{e^9}$ is not an integer, but can we prove this? If you're thinking of asking Wolfram|Alpha, be warned: it gives different answers to the questions "is exp(exp(9)) an integer" (WA says "No") and "is exp(exp(9.0)) an integer" (WA says "Yes"). 

We could prove it is not an integer by providing explicit bounds. If you really wanted to, you could note the following common inequality, holding for all positive $n$:
$$\left(1+\frac{1}n\right)^n<e<\left(1+\frac{1}n\right)^{n+1}.$$
Which yields a series of rational bounds for $e$, where both sides converge to $e$ as $n$ heads to infinity. The point here is that, if we know that $x$ is in the interval $\left[\frac{a}n,\frac{b}{n+1}\right]$, then it follows that
$$\left(1+\frac{1}n\right)^{a}<e^x<\left(1+\frac{1}n\right)^{b}.$$
Taking such an approximation for large enough integer $n$ yields arbitrarily tight rational bounds on $e^x$. From those bounds, you can prove bounds on $e^{e^x}$. I would imagine that you need to find fairly tight bounds, but it's nothing a computer couldn't do.
Now figuring out if it's rational... that's a different story.

Okay, having actually tried to make my computer do the above, it turns out that $e^x$ grows really fast near $x=e^9$ and that the above bounds work in theory, but only slowly. It doesn't quite cut it unless your computer is bigger than mine. We could swap out whatever bounds we want though - for instance, a crude bound arising from the series for $e^x$ is (assuming $x$ is positive and $n\geq 2x$).
$$\left(\sum_{i=0}^n\frac{x^i}{n!}\right)<e^x<\left(\sum_{i=0}^n\frac{x^i}{n!}\right)+\frac{x^n}{n!}$$
converges much faster, and would probably yield the desired bound with some effort. You could also try rounding rationals down (in the lower bound) and up (in the upper bound) to shave some time off doing exact rational arithmetic.
