I was taking a look at the following function: $$f(x) = \ln^{\ln (x)}(x)$$ I see it's not defined in $\mathbb{R}$ for any $x \in (-\infty, 1)$. I understand why it's not defined in ${1}$, as $f(1)$ would be $0^0$.
However, why isn't it defined for $x\in(0,1)$?
If we let $x=\frac{1}{e}$, why don't we have $$f\left(\frac{1}{e}\right) = -1^{-1}= -1$$?
Let $x = \mathrm{10}^{-1/2} = 0.316\dots \text{.}$ Then $$(\log (10^{-1/2}))^{\log (10^{-1/2})} = (-1/2)^{-1/2} = \frac{1}{\sqrt{-1/2}} \text{.} $$ Since we are working in real numbers, we do not take square roots of negative numbers.
This occurs for every $x = 10^{-p/q}$ for $0 < p < q$ and even $q>0$. This removes a dense subset of the interval $(0,1)$ from the domain. There are still some points remaining in the domain in the interval $(0,1)$, but compactly writing down that set is infeasible.