I have seen some elaborate arguments to the effect that if an $ n \times n$ matrix is invertible it has 2-sided inverses. We show a 1-sided inverse implies matrix is ( represents) either injections or surjections (In the case of matrices, implies both.)
But this proof seems simpler:
Assume $AB=I_n \Rightarrow $ A has a right inverse.
Then,
$ABA=I_nA=A \Rightarrow A(BA)=A $ $(*)$ Now
we know if $AB=I_n$ then both $A,B$ are invertible (e.g. $AB=I_n \Rightarrow \det(AB)=\det A \det B=\det I_n=1 $ so $\det A, \det B \neq 0$).
So,
multiplying both sides of $(*)$ by $A^{-1}$:
$A^{-1}(ABA)=(A^{-1}A)(BA)=BA=A^{-1}A=I_n$
Isn't this a simpler and valid proof? (Although I don't see how we are using any special properties of matrices)