Relation between determinant and matrix rank Let $A$ a square matrix with the size of $n \times n$.  
I know that if the rank of the matrix is $<n$, then there must be a "zeroes-line", therefore $\det(A)=0$.

What about $\text{rank}(A)=n$? Why does it imply $\det(A)\ne0$?  

Of course, there is no "zeroes-line", but that doesn't prove it yet.
I've seen a proof in a book which does this conclusion immediately, but IMHO this alone, doesn't prove it.  
What's the missing part?   
 A: Let $A$ be an $n\times n$ matrix. 
Note that $\det(A) \neq 0$ iff the rows are linearly independent iff $rank(A)=n$.
A: The rank of $A$ can be viewed as $m$ where $m$ is the size of the largest non-zero $m\times m$ submatrix with non-zero determinant.
Alternatively, you can row reduce the matrix to give you an upper triangular matrix using row interchanges and adding scalar multiples of a row to another row.  This will only affect the sign of the determinant.  If an $n \times n$ matrix has rank $n$ then it has $n$ pivot columns (and therefore $n$ pivot rows).  This means you will be able to row reduce it to an upper triangular form with pivots along the diagonal.  The determinant is the product of these elements along the diagonal.  Can you prove that?  Pivots are necessarily non-zero and therefore their product is non-zero, regardless of sign.
A: I have an idea: if $$rank(A_{nxn}) = n $$ then $$A$$ must be invertible (proof : https://sharmaeklavya2.github.io/theoremdep/nodes/linear-algebra/matrices/full-rank-inv.html) so $$A^{-1} =  \frac{adj(A)}{det(A)}$$ exists and therefore  $$\rightarrow det(A) \neq 0$$
