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.
