A standard-normal matrix with at least as many columns as rows has full row rank almost surely

For $k=0$, the first column is nonzero almost surely: the event $G_1=0$ is contained in the event that its first coordinate is zero, which has probability zero because a standard normal has a density. For $1\le k<p$, suppose the first $k$ columns are linearly independent. For each fixed such tuple $v_1,\dots ,v_k$, their span $V$ is proper; by [F3] choose $a\mathrm{\ne }0$ with $a\perp V$. The next column is independent of this tuple by [F1], and [F2] gives $a^T{G}_{k+1}\sim N(0,\mathrm{\parallel }a{\mathrm{\parallel }}^2)$, so $\mathbb{P}(G_{k+1}\in V)=0$ because that event is contained in $\{a^T{G}_{k+1}=0\}$. The set of dependent $(k+1)$-tuples is Borel, since dependence is the simultaneous vanishing of its $(k+1)\times (k+1)$ minors and [F5] makes these determinant maps Borel. Integrating these zero-probability sections by [F1] and [F4], and adding the null set of dependent first $k$-tuples, shows that the first $k+1$ columns are linearly independent almost surely.

Applying step 1.1 successively through $k=p-1$ shows that the square matrix $B=[G_1\mathrm{\mid }\cdots \mathrm{\mid }{G}_p]$ is invertible almost surely. For any nonzero $v\in {\mathbb{R}}^p$, the first $p$ coordinates of $G^{T}v$ are $B^{T}v\mathrm{\ne }0$, so $v^{T}G{G}^{T}v=\mathrm{\parallel }{G}^{T}v{\mathrm{\parallel }}^2>0$. Thus $G{G}^T$ is positive definite almost surely.