Show that Col(A)=Col(AB). A and B are square matrices of the same size, B is invertible

I got one exercise that says $A$ and $B$ are both matrices of $n \times n$, $B$ is invertible, show that $Col(A) = Col(AB)$

I am a newb in Maths and here is what I come up with:

$$ \forall \vec x \in \Bbb R^n: A \vec x=\vec v\\ B \vec x=\vec u\\$$ Since B is invertible, its column vectors are linear independent and $\vec u=B\vec x$ spans $\Bbb R^n$, implies that $\vec u = \vec x$.

$\therefore AB \vec x=A(B \vec x)=A\vec u=A\vec x$ then $Col(A) = Col(AB)$

However I am afraid I am wrong and this is not how one would write for proof. Any help on how to prove?

Here's how I would write what you're going for:

Suppose that $v$ is in $col(AB)$. That is, there is some $x$ for which $v=ABx$. We find that if we set $y=Bx$, then $y$ is such that $Ay=v$. This means that $v$ is in $col(A)$.

Suppose on the other hand that $v$ is in $col(A)$. That is, there is some $y$ for which $v=Ay$. Because $B$ is invertible, we can find an $x$ for which $Bx=y$. However, that means that $v=ABx$, which is to say that $v$ is in $col(AB)$.

So, the two column spaces are the same, since they contain the same vectors.

Personally, I would prove it looking at the associated linear maps:

Let $f$ the endomorphism of $K^n$ (I don't know what's your base field) with matrix $A$ in the canonical basis, $g$ the endomorphism with matrix $B$. $\DeclareMathOperator{\col}{Col}\DeclareMathOperator{\im}{Im}\col A$ is simply $\im f$, ans $$\col(AB)=\im(f\circ g)=f\bigl(\im g\bigr)=f(K^n)=\im f=\col A$$ since, as $B$ is invertible, $g$ is an automorphism.