Associativity probably isn't the right word here, but close enough.
Suppose $M$ is a vector space over a field. Suppose $2 \leq \text{dim}(M) \leq \infty$. Suppose $N = \bigwedge^2 M$.
Then I want to see when$\bigwedge^2N \not \cong \bigwedge^4 M$, i.e, $\bigwedge^2(\bigwedge^2M) \not \cong \bigwedge^4 M$, and when $\bigwedge^2 N \cong \bigwedge^4 M$.
Observation: When $2 \leq \text{dim}(M) < \infty$, we have that $\bigwedge^2 N \not \cong \bigwedge^4 M$. This comes from observing the dimensions of the two:
Suppose $\text{dim}(M) = k$. Then if $\{e_1, \cdots, e_k\}$ is the basis for $M$. Then I believe that $\{(e_i, e_j) : 1 \leq i < j \leq k\}$ and $\{(e_i, e_j, e_l, e_m) : 1 \leq i < j < l < m \leq k\}$ is the bases for $\bigwedge^2M$ and $\bigwedge^4 M$ respectively.
Using this idea then: one should come to the conclusion that $\text{dim}(\bigwedge^2N) = {\text{dim}N \choose 2} = {{k \choose 2} \choose 2}$ and $\text{dim}(\bigwedge^4M) = {k \choose 4}$.
From here I believe it's easy enough to show that ${{k \choose 2} \choose 2} > {k \choose 4}$ (for $k \geq 2$) and we're done. $\square$
Question: What about when $\text{dim}(M) = \infty$, i.e, when it's countably infinite. It's unclear to me what happens here- the dimensions match up I believe (you just get $\infty$ for both $\bigwedge^2 N$ and $\bigwedge^4 M$). So is it $\bigwedge^2 N \cong \bigwedge^4 M$? How does one make the bijections between the bases? Do we just say that the cardinality of both bases is $\aleph_0$? How do we show that the bases have this cardinality? I was thinking of some sort of injection like: $(e_i, e_j) \mapsto (e_{2^i3^j})$ and similarly $(e_i, e_j, e_l, e_m) \mapsto (e_{2^i3^j5^l7^m})$? But all of a sudden this has turned into a set theory question which is weird to me. I was wondering if there was a more standard way to prove this?