Group property for division conversion

Question

In a group, based on what property the following is possible

$ (\frac{B}{A})^a = g^{ab - a^2} $

where

$ A = g^a $

$ B = g^b $

$a \epsilon Z_p$

$b \epsilon Z_p$

$Z_p$ is a cyclic group and $g$ is a generator of the group

with reference to first page of paper https://eprint.iacr.org/2015/267.pdf

Answer 1

$(\frac{B}{A})^a=(g^{b-a})^a=g^{ab-a^2}$

I think the main property here is:

$g^ng^m=g^{n+m}$

$g^n/g^m=g^{n-m}$

$(g^n)^k=g^{nk}$

which is similar to the rules for exponential in real numbers.