Proof of Theorem NI

(⇐) Since $A$ is invertible, we can write $I_n=A\inverse{A}$ (Definition MI). Notice that $I_n$ is nonsingular (Theorem NMRRI) so Theorem NPNT implies that $A$ (and $\inverse{A}$) is nonsingular.

(⇒) Suppose now that $A$ is nonsingular. By Theorem CINM we find $B$ so that $AB=I_n$. Then Theorem OSIS tells us that $BA=I_n$. So $B$ is $A$'s inverse, and by construction, $A$ is invertible.