Let $A\in \mathcal{M}_{n,p}(\mathbb{R})$ and $G={}^{t\!}AA$
We assume that $\operatorname{rk}(A)=p$
To show that $\det(G)>0$ the argument provided is $\det G=(\det A)^2>0$, unfortunately I don't understand, because for me this argument holds only if $A\in \mathcal{M}_{p}(\mathbb{R})$ (a square matrix)
The expression $(\det A)^2$ indeed doesn't make sense because $A$ is not necessarily square. As pointed out in a comment, you may use Cauchy-Binet formula to prove the statement. Alternatively, as $A$ has full column rank, if $x\in\mathbb R^p$ is nonzero, then $Ax\ne0$ and $x^TGx=x^TA^TAx=\|Ax\|^2>0$. Therefore $G$ is symmetric positive definite and it has a positive determinant.