If $A \in \mathbb{C}^{m\times n}$ is full-column rank matrix, then is rank($AB$) = rank ($BA$) = rank($B$)?

Supposing $m=k$ so that $BA$ is defined, we can use Sylvester's rank inequality combined with

rank$(AB)\leq$ min$\{$rank$(A)$, rank$(B)\}$. So we have

rank$(A)+$rank$(B)-n \leq$ rank$(AB)\leq$ min$\{$rank$(A)$, rank$(B)\}.$

If $A$ is full column rank then we must have rank$(B) \leq$ rank$(A)$, and we have rank$(A)=n$, so the inequality reduces to

rank$(B) \leq$ rank$(AB)\leq$ rank$(B),$

and we have a similar result for rank$(BA)$, proving the first part.

Note that $N(B) \subseteq N(AB)$ - it is easy to prove this using the standard way of taking $v \in N(B)$ and then showing that we must have $v \in N(AB)$.

You will have the last equality if $A^{-1}$ exists...again this is not difficult to prove - you must just show that $v \in N(AB)$ implies $v \in N(B)$ if $A$ is invertible.

Summary: the equation $\def\rk{\operatorname{rank}}\rk(AB)=\rk(B)$ holds, but (supposing $BA$ is defined) one can have $\rk(BA)\neq\rk(B)$. Simple example of the latter: $$ A=\begin{pmatrix}1\\0\end{pmatrix} ,\qquad B=\begin{pmatrix}0&1\end{pmatrix} ,\qquad BA=(0) ,\qquad\text{( while } AB=\begin{pmatrix}0&1\\0&0\end{pmatrix} \text{).} $$ One does have (with the given hypotheses) $\ker(AB)=\ker(B)$, which more generally holds whenever $\ker(A)\cap\operatorname{range}(B)=\{0\}$ (the given hypotheses say that $\ker(A)=\{0\}$, which is stronger).

Saying that $A$ has full column rank is saying that the columns of $A$ are linearly independent, which means that the map $x\mapsto Ax$ is injective: since $Ax$ is forming a linear combination of the columns of $A$ with the entries of $x$ as coefficients, linear independence says that $Ax=0$ only has the trivial solution $x=0$. Alternative formulation: $\ker(A)=\{0\}$.

Now an injective linear map sends linearly independent sets to linearly independent sets, so a maximal independent set of columns of $B$ corresponds (at the same column indices) to a maximal independent set of columns of$~AB$, which gives $\rk(AB)=\rk(B)$. As the counterexample shows this argumentation does not work when $A$ is the right factor, as in $BA$.

As for $\ker(AB)$, clearly every $x$ with $Bx=0$ also has $ABx=0$, so $\ker(B)\subseteq\ker(AB)$ without any hypotheses on $A$ (other than that $AB$ is defined). For a condition for the reverse inclusion; if we imagine that $Bx\neq0$ but nonetheless $ABx=0$, this means that $Bx$ is a nonzero vector in $\ker(A)$. Clearly when $\ker(A)=\{0\}$ this cannot happen, and more generally it cannot whenever $\ker(A)\cap\operatorname{range}(B)=\{0\}$, since $Bx$ must be a vector of that intersection.