What follows is a greatly expanded version (essentially an excerpt from my personal notes on ordinal arithmetic with operations beyond exponentiation) of my earlier comments about one of the things you bring up in your question. I originally used comments because what I said didn’t really address your question. However, someone has requested that I covert those comments into an answer.
When defining tetration and higher order operations on transfinite ordinal numbers, what you call the right-associative version doesn’t work very well, and the left-associative version is used. To illustrate this problem with transfinite ordinals, I’ll give a summary of some introductory aspects of tetration for transfinite ordinal numbers. For basic definitions and results about addition, multiplication, and exponentiating of ordinal numbers, see this 22 September 2006 sci.math post (25 September 2006 revised version). Incidentally, I planned to continue those posts (see here, for example), but I wound up getting very busy at work (my day job, which had nothing to do with these pursuits), plus I later felt that trying to write all this stuff in ASCII format for posting was too much of a time-sink. At some point I plan to provide a survey of “higher order operations for ordinal numbers” as an answer to some stack exchange question (see my comments to this question), but I don’t know when I’ll get around to doing it (could be several years from now).
Ordinal Tetration. Fix an ordinal $\alpha$. We define $\, \sideset{_{}^\beta}{}\alpha \,$ by transfinite induction on $\beta$ as follows.
(base case) $\;\; \sideset{_{}^0}{}\alpha = 1\;$ and $\; \sideset{_{}^1}{}\alpha = \alpha$
(successor case) $\;\; \sideset{_{}^{\beta + 1}}{}\alpha \; = \; \left(\sideset{_{}^{\beta}}{}\alpha \right)^{\alpha} \;$ for each $\; \beta \geq 1$
(limit case) $\;\; \sideset{_{}^{\lambda}}{}\alpha \; = \; \sup \left\{\sideset{_{}^{\beta}}{}{\alpha}: \; \beta < \lambda \right\}\;$ if $\; \lambda \;$ is a nonzero limit ordinal
Ordinal Tetration vs Usual Tetration. In the case of finite ordinals (i.e. non-negative integers), this is NOT the same as the usual tetration operation. For instance,
$$\sideset{_{}^4}{}\alpha \; = \; \left( \sideset{_{}^3}{}{\alpha} \right)^{\alpha} \; = \; \left( \left( \sideset{_{}^2}{}{\alpha} \right)^{\alpha} \right)^{\alpha} \; = \; \left( \left( {\alpha}^{\alpha} \right)^{\alpha} \right)^{\alpha} \; = \; {\alpha}^{{\alpha}^{3}} $$
In fact, it follows from the result proved further below that
$$\sideset{_{}^{\epsilon_0}}{}{\epsilon_0} \; = \; {\epsilon_0}^{{\epsilon_0}^{\epsilon_0}} $$
Moreover, we also have $\; \epsilon_0 \; = \; \sup\left\{\omega, \; \sideset{_{}^{\omega}}{}{\omega}, \; \sideset{_{}^{\sideset{_{}^{\omega}}{}{\omega}}}{}{\omega}, \; \ldots \right\}$, and hence it follows that
$$ \epsilon_0 \; = \; {\omega}^{{\omega}^{{\omega}^{{\cdot}^{{\cdot}^{\cdot}}}}} \; = \; \sideset{_{}^{\sideset{_{}^{\sideset{_{}^{\sideset{_{}^{\sideset{_{}^{\cdot}}{}{\cdot}}}{}{\cdot}}}{}{\omega}}}{}{\omega}}}{}{\omega} $$
Strong Tetration and Weak Tetration. There does not seem to be a standard term for this distinction in the literature, probably because this distinction does not arise very often. Mark Neyrinck’s May 1995 undergraduate thesis An Investigation of Arithmetic Operations uses the term top-down and bottom-up. I propose, when these two types of tetration are being discussed together, to use the term strong tetration for the ordinary notion of tetration and the term weak tetration for the ordinal notion of tetration as defined above.
Top to Bottom Convention for Finite-Length and $\omega$-Length Towers of Exponentiation and Tetration.
When writing finite-length towers of exponentiation or tetration, I will assume the evaluation is from top to bottom, as is standard convention. Also, the presence of a single ending ellipsis in such a tower represents the supremum of the corresponding finite-length towers that context suggests. Note, however, that when a finite-length TETRATED TOWER of ordinals appears, where we are assuming the evaluation is from top to bottom, the individual tetration operations that are to be performed are those of weak tetration.
Non-Associativity of Exponentiation. The presence of different notions of tetration is due to the non-associativity of exponentiation. For a fixed sequence of FINITE ordinals (i.e. natural numbers), ordinary tetration (evaluate repeated exponentiation from top to bottom) generally results in the greatest value, and the notion we're using (evaluate repeated exponentiation from bottom to top) generally results in the least value, from among all the possible methods of evaluating repeated exponentiation. [For more on this issue, see Gobel/Nederpelt, The number of numerical outcomes of iterated powers, American Mathematical Monthly 78 #10 (December 1971), 1097-1103.] Doner/Tarski’s 1969 paper An extended arithmetic of ordinal numbers explains on p. 113 that there seem to be insurmountable problems in formulating a useful notion of tetration for ordinals in which repeated exponentiation is evaluated from top to bottom.
Why Strong Tetration is Not Useful for Transfinite Ordinals.
The ordinal $\epsilon_0$ is useful in seeing what goes wrong. Suppose we define tetration in the usual way (i.e. strong tetration), so that the successor case is defined as $\; \sideset{_{}^{\beta + 1}}{}\alpha \; = \; {\alpha}^{\left(\sideset{_{}^{\beta}}{}\alpha \right)} \;$ for $\; \beta \geq 1.$ Then under this definition of tetration we would have $\;\sideset{_{}^{2}}{}\omega = {\omega}^{\omega},$ $\; \sideset{_{}^{3}}{}\omega = {\omega}^{{\omega}^{\omega}},$ $\;\sideset{_{}^{4}}{}\omega = {\omega}^{{\omega}^{{\omega}^{{\omega}}}}, \; \dots,$ $\;\sideset{_{}^{\omega}}{}\omega = \epsilon_0.$ [We would keep the limit case of the definition the same, of course.] But now watch what would happen after this: $\; \sideset{_{}^{\omega + 1}}{}\omega = {\omega}^{\left(\sideset{_{}^{\omega}}{}\omega \right)} = {\omega}^{\epsilon_0} = \epsilon_0,\;$ and then $\; \sideset{_{}^{\omega + 2}}{}\omega = {\omega}^{\left(\sideset{_{}^{\omega + 1}}{}\omega \right)} = {\omega}^{\epsilon_0} = \epsilon_0,\;$ and then $\; \sideset{_{}^{\omega + 3}}{}\omega = {\omega}^{\left(\sideset{_{}^{\omega + 2}}{}\omega \right)} = {\omega}^{\epsilon_0} = \epsilon_0,\;$ and so on. Thus, we would have $\; \sideset{_{}^{\omega + n}}{}\omega = \epsilon_0\;$ for each $n < \omega.$ Therefore, since $\; \sideset{_{}^{\omega \cdot 2}}{}\omega = \sup\left\{\sideset{_{}^{\omega + n}}{}\omega: \; n < \omega \right\},\;$ we would then get $\; \sideset{_{}^{\omega \cdot 2}}{}\omega = \epsilon_0.$ In fact, one can show by straightforward transfinite induction that $\; \sideset{_{}^{\beta}}{}\omega = \epsilon_0 \;$ for each $\; \beta \geq \omega$ if strong tetration is used.
Two Formulas for $\;\sideset{_{}^{\beta}}{}\alpha \;$ in Terms of Exponentiation:
$\;$ Let $\alpha \neq 1$ be an ordinal.
- $\;\; n < \omega \;$ implies $\;\; \sideset{_{}^{n+1}}{}\alpha \; = \; {\alpha}^{{\alpha}^{n}} $
- $\;\; \beta \geq \omega \;$ implies $\;\; \sideset{_{}^{\beta}}{}\alpha \; = \; {\alpha}^{{\alpha}^{\beta}} $
One Formula for $\;\sideset{_{}^{\beta}}{}\alpha \;$ is Possible. These two formulas can be replaced with the single formula $\; \sideset{_{}^{1 + \beta}}{}\alpha \; = \; {\alpha}^{{\alpha}^{\beta}},$ but I prefer putting the result into two separate formulas that are more directly comprehended. The two separate formulas are more directly comprehended because the reader does not have to mentally supply the result "$1 + \beta = \beta$ for $\beta \geq \omega$" to get the simpler version when $\beta$ is infinite. Also, I think there is potential for error when using $\; \sideset{_{}^{1 + \beta}}{}\alpha \; = \; {\alpha}^{{\alpha}^{\beta}},$ because the reader might mistakenly think $1 + \beta$ is a typo and that the writer had intended $\beta + 1$ instead.
Proof of First Formula. We use mathematical induction on $n < \omega$. (base case) The result for $n = 0$ follows from $\sideset{_{}^1}{}\alpha = \alpha$ and ${\alpha}^{{\alpha}^0} ={\alpha}^1 = \alpha$. (successor case) Assume the result holds for $n = k.$ Then we have $\sideset{_{}^{k+1}}{}\alpha = \left(\sideset{_{}^{k}}{}\alpha \right)^{\alpha} = \left( {\alpha}^{{\alpha}^k} \right)^{\alpha} = {\alpha}^{{\alpha}^{k} \cdot \alpha} = {\alpha}^{{\alpha}^{k+1}}$ (the induction hypothesis is used in the 2nd equality), which shows that the result holds for $n = k+1.$
Proof of Second Formula. We use transfinite induction for ordinals $\beta \geq \omega$. (base case) We show the result is true for $\beta = \omega.$ By definition, we have $\sideset{_{}^{\omega}}{}{\alpha} = \sup\left\{\sideset{_{}^{n}}{}{\alpha}: \; n < \omega \right\}$ which, by making use of monotonicity in the tetrated exponent, equals $\sup\left\{\sideset{_{}^{n+1}}{}{\alpha}: \; n < \omega \right\}.$ Using what we just proved in #1, this is equal to $\sup \left\{{\alpha}^{{\alpha}^{n}}: \; n < \omega\right\} = {\alpha}^{{\alpha}^{\omega}}.$ The last equality follows by an application of continuity in the exponent. (successor case) Assume the result holds for $\beta = \eta,$ where $\eta \geq \omega$ is fixed. Then we have $\sideset{_{}^{\eta + 1}}{}\alpha = \left(\sideset{_{}^{\eta}}{}\alpha \right)^{\alpha} = \left( {\alpha}^{{\alpha}^{\eta}} \right)^{\alpha} = {\alpha}^{{\alpha}^{\eta} \cdot \alpha} = {\alpha}^{{\alpha}^{\eta + 1}}$ (the induction hypothesis is used in the 2nd equality), which shows that the result holds for $\beta = \eta + 1.$ (limit case) Let $\lambda > \omega$ be a limit ordinal and assume the result holds for all $\beta$ such that $\omega \leq \beta < \lambda.$ Then we have $\sideset{_{}^{\lambda}}{}\alpha \; = \; \sup \left\{\sideset{_{}^{\beta}}{}{\alpha}: \; \omega \leq \beta < \lambda \right\} \; = \; \sup \left\{{\alpha}^{{\alpha}^{\beta}}: \; \omega \leq \beta < \lambda \right\} \; = \; {\alpha}^{{\alpha}^{\lambda}}$ (the induction hypothesis is used in the 2nd equality), which shows that the result holds for $\beta = \lambda.$
