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.
Generally, as explained in this answer, in any polynomial expression of integers, i.e, an expression composed of sum and product operations on integers, by inductively applying the congruence sum and product rules we obtain a congruent expression if we replace the arguments of sums and products by congruent integers. Above is a special case: replace all $\,a\bmod m\,$ by $\,a$.
