Generally, to prove an equality about mod as an operation it is usually easiest to convert it into an equivalent but more flexible congruence relation form, via the following equivalence (cf. here)
$$ (g\bmod m) = (h\bmod m)\color{#c00}\iff g\equiv h\!\!\!\pmod {\!m}\qquad $$
$\begin{align}
{\rm so}\ \ \ (g^b\bmod m)^a\bmod m &= (g^a \bmod m)^b\bmod m\\[.2em]
\color{#c00}\iff (g^b\bmod m)^a &\equiv (g^a \bmod m)^b \!\!\!\pmod{\!m}\\[.2em]
\iff (g^b)^a &\equiv (g^a)^b\!\!\! \pmod{\!m}\ \ {\rm by}\ \ \color{#0a0}{\rm CPR}\ \ \color{#c00}{\&}\ \ g^{n}\bmod m\equiv g^n\!\!\!\!\pmod{\!m}
\end{align}$
that's true by $\,(g^b)^a = g^{ab} = (g^a)^b\,$ by integer power laws. Here $\color{#0a0}{\rm CPR} = $ Congruence Power Rule
Key Idea $ $ Generally, as above, using the above equivalence and congruence laws, we can erase all $\!\bmod m$ operations in a polynomial expression (i.e. where mod appears in arguments of sums & products, but not in exponents) to obtain an equivalent congruence, e.g. for the above
$$\begin{align}(\color{#c00}{g^b\color{#bbb}{\bmod m})^a\color{#bbb}{\bmod m}} \,&=\, \color{#0a0}{(g^a \color{#bbb}{\bmod m})^b\color{#bbb}{\bmod m}}\\[.2em]
\iff \color{#c00}{(g^b)^a} &\equiv\, \color{#0a0}{(g^a)^b}\!\!\pmod{\!m}
\end{align}\qquad$$
Since congruence arithmetic inherits all common (ring) arithmetic laws (commutative, associative, distributive), it is much easier to use our well-honed arithmetical (vs. divisibility) intuition to prove congruence equations than it is to prove the equivalent divisibility statements.
See also this answer.
