Post History

The power operation is considered right-associative, because the left-associative interpretation would just be a more complicated notation for the product of the exponents:

$a^{bc} = (a^b)^c \neq a^{b^c} = a^{(b^c)}$

Similar arguments lead to the same interpretation for higher towers, where it is even more apparent that this interpretation avoids redundancy.

$a^{bc...xy} = ((...(a^b)^c...)^x)^y\neq a^{b^{c^{⋰^{x^y}}}} = a^{(b^{(c^{⋰^{(x^y)}⋱})})}$

I could not quickly find a reference introducing the notation, the authors seem to use this notation without introducing it explicitly.
However the following sources make the intended interpretation very clear with the corresponding definitions for [Knuth's up-arrow notation [Wikipedia]](https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation).


[Knuth, Donald E. (1976). "Mathematics and Computer Science: Coping with Finiteness". Science. 194 (4271): 1235–1242.](http://www.sciacchitano.it/Spazio/Coping%20with%20Finiteness.pdf)

Knuth introduced the arrow notation and illustrates its use with powers of 10:

 > Similarly we can define a number I shall write as $x\uparrow n$ [...] For example $10\uparrow 10$ [...] is usually written as $10^{10}$

Knuth defines the $\uparrow\uparrow$ as $\uparrow$ nested from the right:

 > $x\uparrow\uparrow n = x\uparrow(x\uparrow(...\uparrow x)...))$

With the examples showing the correspondance between arrow and superscript notation it is clear that he also intends the same order of application for a power tower.

> $10\uparrow\uparrow{\color{red}10} = 10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}$

where I inserted the the red ${\color{red}10}$ as I think it is a typo in the source.

See also
[Galidakis, Ioannis and Weisstein, Eric W. "Power Tower." From MathWorld--A Wolfram Web Resource.](https://mathworld.wolfram.com/PowerTower.html)

Equation (1) links the repeated exponentiation notation to the double arrow notation. If we set $n=2$ in equations (2) to (4) we get

$\underbrace{a^{a^{⋰^{a}}}}_{k} $
$ = a\uparrow\uparrow k $ 
$ = a\uparrow\[a\uparrow\uparrow (k-1)\] $ 

$ = a^{\Big(\raise 4pt {\[ {\underbrace{a^{a^{⋰^{a}}}}_{(k-1)}} \]}\Big)}$ 

Applying this definition recursively will nest the rightmost power into the innermost parenthesis.