Show that if $q$ is an odd prime then $3$ divides $\binom{2^q}{2}-1$ part 1

Question

I would like to show that the following is true

If $q$ is an odd prime then $$3 \text{ }\Bigg | \binom{2^q}{2}-1$$

I know that $(3,q) =1$ for $q \neq 3$ and I should be using the following result from Euler $3^{\phi(q)}\equiv 1 \text{ (mod $q$)}$. But I am not certain of the next step.

Answer 1

$$\begin{pmatrix} 2^q \\ 2 \end{pmatrix}-1=2^{q-1}(2^q-1)-1$$

Hence taking $\mod 3$,

$$\begin{pmatrix} 2^q \\ 2 \end{pmatrix}-1 \equiv (-1)^{q-1}((-1)^q-1)-1 \equiv (-1)^q-1-1=-1-2 \equiv 0 \mod 3$$

where I have used the fact that $(-1)^{q} =-1$ since $q$ is odd