Fermat's Library tweeted five hours ago that $e^\pi$ can be proved to be transcendental as follows. Let $A=-1$ and $B=-i$. Then $A$ and $B$ are algebraic numbers, $A\ne0,1$ and $B$ is not rational. Therefore, by Gelfond-Schneider theorem, $$A^B=e^\pi=(e^{i\pi})^{-i}=(-1)^{-i}\tag{1}$$ is a transcendental number.
The same equality $(1)$ is also mentioned in Wikipedia, but I am confused because an answer on this site stated that the rule $x^{yz}=(x^y)^z$ is valid only when $x>0$ and $y$ is real. Does the rule really apply here? Is the proof in the aforementioned tweet valid?
Just because an identity does not hold generically, does not mean that there are no coincidences where it seems to work anyway.
Having chosen a branch of the logarithm, we may write the complex exponential $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{-\mathrm{i} \ln(-1)} $$ and proceed from there.
If we do not wish to choose a branch of the logarithm, we instead write $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{-\mathrm{i} (\ln(-1) + 2 \pi \mathrm{i} k)} $$ for some choice of integer $k$ (which choice is equivalent to a choice of branch). Then we can evaluate $\ln(-1)$ on any branch, for instance $\ln(-1) = \mathrm{i} \pi$. And finally, \begin{align*} -\mathrm{i} (\ln(-1) + 2 \pi \mathrm{i} k) &= -\mathrm{i}(\mathrm{i} \pi + 2\pi \mathrm{i}k) \\ &= \pi + 2 \pi k \text{.} \end{align*} So $$ (-1)^{-\mathrm{i}} = \mathrm{e}^{\pi + 2 \pi k} \text{,} $$ where the choices of $k$ correspond to different choices of branch of logarithm in defining the complex exponential. If we choose the branch with $k = 0$, we recover the claimed identity.