Method for calculating exponents and beyond with logarithmic growth provided a memorized set of smaller problems

Question

This question is very hard to word so I'm sorry about that, but here goes a try.

With Addition

Let's assume I have all addition facts from 1-10 memorized. When doing $125+126$ i will employ these steps

1. $5+6 = 11$
2. $2 + 2 + 1 = 5$ carrying the one from above
3. $1 + 1 = 2$

So it is $251$

Now if I multiply both of the previous numbers by 2, my step number stays the same $250 + 252$

1. $2 + 0 = 2$
2. $5 + 5 = 10$
3. $2 + 2 + 1 = 5$ so it's $502$.

In general, the number of operations in my context taken to perform an addition problem assuming the 1-10 addition facts are memorized with operands below size $n$ takes $\log(n)+1$ steps and thus grows very slowly.

With Multiplication

I am not worrying about the carrying as it still is logarithmic even if you count carrying as a step and it just makes explaining more complex

Assuming we have again facts 1-10 memorized, and we can multiply by 10 easily lets do $125 \times 126$

1. $6 \times 5 = 30$
2. $6 \times 2 = 12$
3. $6 \times 1 = 6$
4. $2 \times 5 = 10$
5. $2 \times 2 = 4$
6. $2 \times 1 = 1$
7. $1 \times 5 = 5$
8. $1 \times 2 = 2$
9. $1 \times 1 = 1$

and then we do all the easy carrying and multiplying by 10 that every school child knows.

Now for multiplication we generally take less than or equal to $\log(n)^2$ steps where both operands are less than or equal to n.

With Exponentation

1. Let us take $125^{126}$ is there a way to solve this and any other exponent problem provided I have some finite set of problems memorized (assuming this is 1-10, but it doesn't really matter) that takes a logarithmic number of steps based on input number size?
2. If not can it be proven why not?

Generally

Assuming I have a function called $f$ like

$$f(n,b)(a)=\begin{cases} a+b&&\text{if}~ n=1 \\ (f(n,a))^{\circ b-1}(a)&&\text{otherwise} \end{cases}$$

By the way, I'm using this notation for calling a function repeatedly, and also I don't know if currying works like this normally in math (I'm a programmer). Just to clear things up here is the same function in python

def f(n,a,b):
  if n == 1:
    return a+b
res = f(n-1,a,a)
  for i in range(0,b-2):
    res = f(n-1,res,a)
  return res

anyway, this function behaves so that

$f(1,b)(a)$ is equivelent to $a + b$

$f(2,b)(a)$ is equivelent to $a \times b$

$f(3,b)(a)$ is equivelent to $a ^ b$

and so on.

3. So assuming we have any given integer for the $n$ parameter is there a way to tell if we can simplify this problem by requiring a logarithmic amount of steps.
4. If not, can it be proven.

Conclusion

I would be happy if any of questions 1-4 are answered all though I would be delighted to see them all answered!

I'm open to any ways to make this easier to understand. Thanks!