On an issue

A question:
for a given a, b seek

$$a^{a^{a^{a^{.^{.^a}}}}} \pmod p$$ (Which has a b a).

For this problem, there are two methods to understand.
1. If it is calculated from the bottom up, so it's actually a formula:

$$a^{a^{b-1}}$$
Then we index above formula can be used by Fermat's Little Theorem $mod$ $p-1$ To get a smaller index, and then do a quick power can be.
Complex as $O(log_2p+log_2(p-1))$
2. If it is calculated from the top downwards, it can be extended by Euler's theorem:
$$a^m \equiv a^{min(m,(m\ mod\ \phi(p))+\phi(p))} \pmod p$$
Recursive index continued until $\phi(p)=1$ When, Back 1.
Operators Euler function recursively.