Proof - Square Matrix has maximal rank if and only if it is invertible
Suppose $A\in F^{n \times n}$. If A is invertible then there is a matrix B such that $AB=I$ so the standard basis $e_i$ (the columns of I) is in the image of A (these vectors are just the image Av where v are the columns of B) - this shows that $\dim (Im(A)) = n$.
On the other hand, if $\dim (Im (A))=n$ then for every i there is $v_i$ such that $A v_i = e_i$. Let B be the matrix with columns $v_i$ then $AB=I$ and A is invertible.