$M$ is a vector space, and $N = \bigwedge^2 M$ . If $2 \leq \dim(M) < \infty$ then is $\bigwedge^2 N \not \cong \bigwedge^4 M$

Question

Suppose $F$ is some base field and $M$ is a $F$-vector space. Then let $N = \bigwedge^2 M$.

Now suppose that $2 \leq \dim(M) < \infty$- then I want to see that $\bigwedge^2 N \not \cong \bigwedge^4 M$

Not sure about this- intuitively it seems like I should get something like $\dim(\bigwedge^2N) = 4 \times \dim(M)^2$ and $\dim(\bigwedge^4M) = 4 \times \dim(M)$ and these two things aren't equal when $\dim(M) > 1$.

I think that's about it, but I was wondering how I would do it formally.

Also, I was curious about $\dim(M)$ is countably infinite. In this case, I think we should get $\bigwedge^2 N \cong \bigwedge^4 M$, but again I'm a bit confused about formally proving it.