How symmetric is the hierarchy of indecomposable ordinals?
The process of ordinal arithmetic follows a repetitive, inductive pattern from addition to multiplication to exponentiation etc., in which the morphism from one operation to the next looks pretty much the same - in the respect that every step up is an induction beyond the transinfinite limit of the previous operation. I'm calling this a symmetry.
We have the concepts of additively indecomposable ordinals, and beyond that, multiplicatively indecomposable ordinals etc. Does this hierarchy enjoy the same inductive symmetry infinitely far in the same way - i.e. to exponentiation, tetration and beyond?
$1,\omega^\beta$ are additively indecomposable (for all ordinals $\beta$).
$2,\omega^{\omega^\beta}$ are multiplicatively indecomposable.
How does this list continue and what does it look like?
It looks like $0^0=1$ and $1^1=1$ but $(n-1)^{n-1}>n\forall n>2\implies $ so the first two exponentially indecomposable ordinals would be:
$2,\omega$
Is it $2,\omega,\omega^{\omega^{\omega^\beta}}$ for all ordinals $\beta$ and so on? Or does the list of special cases at the start get ever more complicated, and do the power towers just get taller or do they change? Tetration is a complicated matter (at least over regular numbers) as the orders of exponents matter and the rule to always apply powers from the right isn't necessarily well-justified, so I imagine there's the potential to get into considering Dyck words etc. How messy does it get?
I suggest you look into this preprint of Alec Rhea: The Ordinals as a Consummate Abstraction of Number Systems. He studies a hierarchy of such binary operations as sum, product and so on (under the name of "hyperoperation sequence")
As for exponentially indecomposable ordinals defined as ordinals stable under exponentiation, they are $2$, $\omega$, and the solutions of $\omega^{\varepsilon}=\varepsilon$, i.e. so called $\varepsilon$-numbers.