Prove $t^{1\over t-1}$ is rational only when $t=2$ where $t$ is any real number.
This problem arises from me watching the blackpenredpen video solving $x^y=y^x$ where $x\neq y$. https://www.youtube.com/watch?v=PI1NeGtJo7s, in which he gives the solution $x=t^{1\over t-1}$.
First I considered the integer solutions where $x\neq y$ other than $2^4=4^2$ but realized it is trivial because $1<t^{1\over t-1}<2$ for any $t\geq 2$ and $-1>t^{1\over t-1}>-2$ for any $t\leq-2$.
The next question is then, are there any rational solutions other than the $2,4$ pair. It does not seem trivial to me as inequalities do not help.
My guess is that $t^{1\over t-1}$ cannot be rational other than when $t=2$ but I wan't able to prove.
The range of $f(t)=t^{1/(t-1)}$ is $(1,\infty)$ and the function is continuous and monotonic, hence invertible. So there is an infinite (but countable) number of solutions of
$$t^{1/(t-1)}=q$$ where $q$ is rational.
$$t=-\frac1{\log q}W\left(-\frac{\log q}q\right).$$