The problem: Let $A\in\mathbb{K}^{m\times n}$ be a matrix. Show that $\text{rank}({A})$ is identical to the largest number $k\in\{1,2\dots,\min(m,n)\}$ such that the subdeterminant with $i_1<i_2<\dots<i_k\in\{1,\dots,m\}$ and $j_1<j_2<\dots<j_k\in\{1,\dots,n\}$ \begin{equation} \begin{vmatrix} a_{i_1j_1} & a_{i_1j_2} & \dots & a_{i_1j_k}\\ a_{i_2j_1} & a_{i_2j_2} & \dots & a_{i_2j_k}\\ \vdots & \vdots &\ddots &\vdots\\ a_{i_kj_1} & a_{i_kj_2} & \dots & a_{i_kj_k} \end{vmatrix}\neq 0. \end{equation}
My proof attempt: Let $\text{rank}(A)=k$. Now I show that the set of indices is maximal.
Suppose $\exists i_1<\dots<i_{k+1}\in\{1,\dots,m\}$ and $\exists j_1<\dots<j_{k+1}\in\{1,\dots,n\}$ such that \begin{equation} \begin{vmatrix} a_{i_1j_1} & a_{i_1j_2} & \dots & a_{i_1j_{k+1}}\\ a_{i_2j_1} & a_{i_2j_2} & \dots & a_{i_2j_{k+1}}\\ \vdots & \vdots &\ddots &\vdots\\ a_{i_{k+1}j_1} & a_{i_{k+1}j_2} & \dots & a_{i_{k+1}j_{k+1}} \end{vmatrix}\neq 0. \end{equation}
Since the determinant is not equal to zero, the vectors $\begin{pmatrix} a_{i_1j_1}\\ \vdots\\ a_{i_{k+1}j_1} \end{pmatrix},\dots, \begin{pmatrix} a_{i_1j_{k+1}}\\ \vdots\\ a_{i_{k+1}j_{k+1}} \end{pmatrix}$ are linearly independent. This implies that the column vectors $\begin{pmatrix} a_{1j_1}\\ \vdots\\ a_{mj_1} \end{pmatrix},\dots, \begin{pmatrix} a_{1j_{k+1}}\\ \vdots\\ a_{mj_{k+1}} \end{pmatrix}$ of the matrix $A$ are also linearly independent. By definition of the rank of a matrix we can conclude that $\text{rank}(A)\geq k+1>k=\text{rank}(A)$, which is a contradiction. Therefore the set of indices was maximal and $k$ fulfills the requirement. $\blacksquare$
Is this reasoning correct?