Tetration is defined as “repeated exponentiation” - that is, $2$ tetrated to $5$ is equal to $2^{2^{2^{2^2}}}$, just as $2$ exponentiated to $5$ is equal to $2\cdot{2}\cdot{2}\cdot{2}\cdot{2}$ (repeated multiplication). My first basic question is:
Does tetration go “up” or “down”?
That is, stated more clearly, do you resolve the exponentiations on the bottom or the top first? For example, you can solve $2$ tetrated to $5$ as
$$2^{2^{2^{2^2}}} = 4^{2^{2^2}} = 16^{2^2} =256^2=65536 = 2^{16}$$
or
$$2^{2^{2^{2^2}}} = 2^{2^{2^4}} = 2^{2^{16}}= 2^{65536}.$$
The simple fact that we get two different results depending on the order in which we perform the exponentiation makes tetration unique. Addition and multiplication are both commutative (e.g. $2 \cdot (3 \cdot 4) = (2 \cdot 3) \cdot 4$ but exponentiation is not (e.g. $(3^3)^3 \neq 3^{(3^3)}$). This is why we get this ambiguity about tetration that is absent for addition, multiplication, and exponentiation. Another question:
Is this ambiguity present for the further hyperoperations? If so, do they go “up” or “down” as well?
In addition, there is a multiplication identity for exponentiation:
$$(x^m)^n = x^{mn}$$
Is there an analogue to this for tetration or the other hyperoperators?
Finally:
Is there a standard notation for all the hyperoperators? What about hyperoperator roots (just as $\sqrt{x} \cdot \sqrt{x} = x$, a “tetrational square root” defined as $k =$ [$x$ tetrated to $\tfrac{1}{2}$] where $k^k = x$)?
Thanks for your assistance!
