If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$,
thus $\det(A) =0$.
I proved that way, but looks like too simple so I think maybe there is a trick that I missed.
Or I proved it right?
Modification:
($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$
($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)
Now I think I complete the proof.
Best Answer
Perhaps the simplest way to prove this, and in two lines, is the following... but it depends on what definitions you have (all the time square matrices of the same order):
A matrix $\,A\,$ is invertible iff there exists a matrix $\,B\,$ s.t. $\;AB= BA=I\;$ iff when reduced (by rows or columns) no row (column) is all zeros iff $\,\det A\neq 0\,$ .
The last double implication follows from the fact that the determinant of an upper (lower) triangular matrix is the product of the entries in its main diagonal (and, of course, on the fact that interchanging rows only multiplies the matrix by $\,\pm 1\,$)
If your definitions cover the above there you have an easy proof of what you want.