This question is inspired by a challenge on codegolf.stackexchange.com, where the task is to:
Output consecutive digits of $x = \pi^{1/\pi}$ forever.
That is, calculate digits of $x$, and output one by one forever (in theory). As this should go on forever, you must continuously calculate new digits of $\pi$ and $x$. There are known ways to calculate individual digits of $\pi$, but as far as I can tell, digit $n$ of $\pi^{1/\pi}$ may rely on digit $n+m$ of $\pi$.
At some point, you'll encounter something like $...31999999999999999...$ if you calculate $x$ with a given accuracy for $pi$. However, if you had a higher accuracy, this might result in $...3200000000000...$ instead.
I'm fairly certain that it can be proven/shown that digit $n$ may rely on digit $n+m$ of $\pi$ for some $n$ and $m$ (but I can't). What I'm not sure of is, can an upper bound of $m$ be proven? Is it possible to say, with absolute certainty, that the $n$th digit of $x$ is correct if we have calculated for instance $n+10$ digits of $\pi$?
If you do come across $\ldots3199999999\ldots$, then you just have to keep going until the issue resolves itself as either $x\le\ldots3199999999$ or $x\ge\ldots32$. If $x$ is known to be irrational, then you can be sure that this procedure will eventually terminate.
It can be difficult to predict in advance how much look-ahead you will need. In practice, however, unless $x$ has been specially chosen to be awkward, you will only need to calculate a few more decimal places.
But if $x$ is not known to be irrational, then you have a genuine problem on your hands. I am pretty sure that $\pi^{1/\pi}$ is irrational, but I wouldn't know how to prove it.