Let $A$ and $B$ be $n \times n$ matrices. If $AB = I$, then $BA = I$.

Question

Here is Prob. 15, Sec. 2.2, in the book Linear Algebra With Applications by Steven J. Leon and Lisette de Pillis, tenth edition:

Let $A$ and $B$ be $n \times n$ matrices. Prove that if $AB = I$, then $BA = I$. ...

Although this question has several answers here on Math Stack Exchange, I feel that my answer below is more elementary since it uses minimal background knowledge.

My Attempt:

Since $AB = I$, where $I$ is the $n \times n$ identity matrix, we have $$ \det (A) \det(B) = \det (AB) = \det(I) = 1, \tag{1} $$ and hence $$ \det ( B) \neq 0, $$ which shows that $B$ is a non-singular matrix.

Therefore we have $$ B = BI = B(AB) = (BA)B, $$ which implies that $$ B - (BA)B = O, $$ where $O$ is the $n \times n$ zero matrix. The preceding relation can be rewritten as $$ IB - (BA)B = O, $$ and hence $$ (I - BA)B = O, $$ and since $B$ is non-singular, we have $$ (I - BA)BB^{-1} = OB^{-1}, $$ that is, $$ (I - BA)I = O, $$ or in other words, $$ I - BA = O, $$ and hence $$ I = BA, $$ which is the same as $$ BA = I, $$ as required.

Alternatively, from (1) above we can conclude that $$ \det(A) \neq 0 \qquad \mbox{ and } \qquad \det(B) \neq 0, $$ which shows that both $A$ and $B$ are non-singular matrices. Then we obtain $$ A \left( A^{-1} - B \right) = AA^{-1} - AB = I- I = O, $$ and hence $$ A \left( A^{-1} - B \right) = O, $$ which implies $$ A^{-1}A \left( A^{-1} - B \right) = A^{-1} O, $$ or in other words, $$ I \left( A^{-1} - B \right) = O, $$ and hence $$ A^{-1} - B = O, $$ which implies $$ A^{-1} = B, $$ and so $$ BA = A^{-1} A = I, $$ as required.

Is this approach correct and clear enough? If so, then is this proof not simpler than that requiring a knowledge of vector spaces and linear transformations?