My Proof: If $B$ is a matrix that is the inverse of $A$ then $BA=I$, we have that rowspace of $BA$ is a subset of rowspace of $A$. Since $\operatorname{rank}(I_n)=n$ then $\operatorname{rank}(A)\ge n$. But since $A$ is a $n\times n$ matrix then $\operatorname{rank}(A)\le n$. This implies $\operatorname{rank}(A)=n$.
Is this proof ok? The result that rowspace of $BA$ is contained in rowspace of $A$ is something I have proved before this.