Minor of a product is a linear combination

Let $$A\in \mathcal{M_{m,n}}$$ and $$B\in\mathcal M_{n,s}$$ be matrices.Then any minor of order $$k ,1\leq k\leq \min(m,s)$$ of the product $$AB$$ it can be written as a linear combination of minors of order $$k$$ of matrix $$A$$ (or $$B$$).

This one of the theorems in my book. My problem with this theorem is the following

If $$n\leq m\leq s$$ maybe there exists a minor of $$AB$$ of order $$k_0$$ s.t $$n\leq k_0\leq m\leq s$$. The theorem told me this minor can be write as a linear combination of minors of order $$k_0$$ of matrix $$A$$, but all minors of $$A$$ have order less than $$\min(m,n)\leq n \leq k_0$$, so $$A$$ does not have minors of order $$k_0$$ How is this possible?

If all minors of $$A$$ have order less than $$k_0$$, then the theorem is claiming that any order-$$k_0$$ minor of $$AB$$ is a linear combination of nothing. This is simply saying that such a minor must be $$0$$ (because the only linear combination of nothing is $$0$$). And this is indeed true.
Incidentally, the theorem can be made a lot more explicit: Any order-$$k$$ minor of $$AB$$ can be written as a sum of products of the form $$\alpha \beta$$ where $$\alpha$$ is an order-$$k$$ minor of $$A$$ and where $$\beta$$ is an order-$$k$$ minor of $$B$$. See Cauchy-Binet formula: general form for this fact.