Columns of $AB$ are independent $\rightarrow$ Columns of B are independent Let $A\in M_{mxn}(F)$ and $B\in M_{nxk}(F)$

Columns of $AB$ are independent $\rightarrow$ Rank(AB)=k on the other hand $Rank(AB)\leq min(Rank(A),Rank(B))$ therefore $k \leq min(Rank(A),Rank(B))$ due to $B\in M_{nxk}(F)$ $k \leq Rank(B)=k$ and therefore $B$ columns are independent
is it a valid proof? Should I use the Rank?

The proof is correct. But you could also have a different view on this. Let $B=(b_1,\dots, b_k)$ with $b_i$ the columns of $B$; then $AB=(Ab_1,\dots,Ab_k)$, that is, the columns of $AB$ are the vectors $Ab_i$. Now, if there is a non-trivial relation $\sum \lambda_ib_i=0$ (that is, some $\lambda_i\ne 0$) and the $b_i$ are linearely dependent then so are the vectors $Ab_i$ as they satisfy the same relation: $$0=A0=A(\sum \lambda_i b_i)=\sum A(\lambda_i b_i)=\sum \lambda_i Ab_i$$
and so the vectors $Ab_i$ are also linearely dependent.
