Note 0: I use 'minor' to refer to both sub-matrix and its determinant. So when you delete row k and column $l$, you get a minor matrix. Its determinant is a minor determinant.
Is this true for $k=0$ instead of $k=1, ..., \min\{m,n\}$?
Proposition: Let $F$ be a field. For any $A \in F^{m \times n}$ with $1 \le rank(A)=k \le \min\{m,n\}$, we have that some minor determinant $\det(M_{(i,j)})$ of a minor matrix of $M_{(i,j)}$, of size $k \times k$, is nonzero.
I was thinking vacuously yes, but I think it's still no because the empty sum is zero or something (unlike the empty product which is 1). But I think it should be yes because there's no such thing as a $0 \times 0$ matrix...
Note 1: see the comments turned into cw answer
Note 2: I know rank 0 iff the zero matrix. I don't really see how this helps though. My concern is (mainly?) about how you say determinant of nothing is 1 or at least nonzero instead of 0. In this case, refer to Note 1.
If a matrix has rank 0, then it is the zero matrix.