Use ranks, the ranks of $B$ and $BB^\intercal$ are the same; also, the column span is the image vector space, so you will have that $\mathsf{im}(BB^\intercal) \subset \mathsf{im}(B)$ and that the two vector spaces have the same dimensions.
To prove that $\mathsf{rk}(BB^\intercal) = \mathsf{rk}(B),$ consider $x$ such that $BB^\intercal x = 0,$ then $x^\intercal BB^\intercal x = 0,$ or $\mathsf{norm}(B^\intercal x)^2 = 0,$ so $B^\intercal x = 0$ as well. Therefore $\mathsf{ker}(BB^\intercal) \subset \mathsf{ker}(B^\intercal).$ Therefore, the rank of $BB^\intercal$ is $\leq \mathsf{rk}(B)$ and the nullity of $BB^\intercal$ is also $\leq \mathsf{nul}(B^\intercal) = \mathsf{nul}(B),$ so they each must coincide with the corresponding number, given that $\mathsf{rk}+\mathsf{nul}$ must equal the dimension of the vector space.