Please verify this statement and proof. Is it correct?
Here, 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.
Proposition. Let $F$ be a field. Let $n \ge 2$ be an integer. For any $A \in F^{n \times n}$ with $rank(A)=k = n-1$, 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.
Proof:
Step 0.1: Let $k,m,s \ge 1$ be integers. Recall that for any matrix $B \in F^{m \times s}$, we have that $rank(B)=k \le \min\{m,s\}$ if and only if $k$ is the maximal number of $F$-linearly independent rows in $B$ if and only if $k$ is the maximal number of $F$-linearly independent columns in $B$.
Step 0.2: I let all 'vectors' here be column vectors. Let the rows and columns of $A$ be, resp, $a_1^T, ..., a_n^T \in F^{n \times 1}$ and then $b_1, ..., b_n \in F^{n \times 1}$.
Explicitly, $A=\begin{bmatrix} a_1^T\\ a_2^T\\ \vdots\\ a_n^T \end{bmatrix} = \begin{bmatrix} b_1 & b_2 & \cdots & b_n \end{bmatrix}$
Step 1: For $rank(A)=k=n-1$, we have that $A$ has $k$ $F$-linearly independent rows, by Step 0.1. Form a $k \times n$ (sub-)matrix $M$ of $A$ out of those $k$ rows. Specifically, for the rows $a_{1}^T, ..., a_{n}^T$, we use all of them except for exactly 1 $a_s^T$ to form $M$.
Explicitly, $M=\begin{bmatrix} a_1^T\\ a_2^T\\ \vdots\\ a_{s-1}^T\\ a_{s+1}^T\\ \vdots\\ a_n^T \end{bmatrix}$.
Then $M$ has $rank(M)=k$ too.
Step 2: Then $M$ in Step 1 has $k$ $F$-linearly independent columns, by Step 0.1. Form a $k \times k$ (sub-)matrix $L$ of $M$ out of those $k$ columns. Specifically, for the columns $b_{1}, ..., b_{n} \in F^{n \times 1}$ of $A$, we have that $M$ is made up of columns $b^M_{1}, ..., b^M_{n} \in F^{(n-1) \times 1}$, where each $b^M_j$ is the original $b_j$ but without its $s$th entry.
Explicitly, $M=\begin{bmatrix} b^M_1 & b^M_2 & \cdots & b^M_n \end{bmatrix}$.
Then we use all of $b^M_{1}, ..., b^M_{n}$ except for exactly 1 $b_t$ to form $L$.
Explicitly, $L=\begin{bmatrix} b^M_1 & b^M_2 & \cdots & b^M_{t-1} & b^M_{t+1} & \cdots & b^M_n \end{bmatrix}$.
Then $L$ has $rank(L) = k$ (too).
Step 3: But, this time, $L$ in Step 2 is square and has full rank.
Step 4: By invertible if and only if full rank and by Step 3, we have that $L$ is invertible.
Step 5 By invertible if and only if non-zero determinant and by Step 4, we have that $L$ has nonzero determinant.
Step 6: As for the required minor determinant $\det(M_{(i,j)})$, choose the $(i,j)$ that gives $L$ as the minor matrix $M_{(i,j)} = L$. This is $(i,j)=(s,t)$.
QED