Cartesian Product is not Associative

Theorem
Let $A, B, C$ be non-empty sets.

Then:
 * $A \times \left({B \times C}\right) \ne \left({A \times B}\right) \times C$

where $A \times B$ is the cartesian product of $A$ and $B$.

Proof
By definition:
 * $A \times B = \left\{{\left({a, b}\right): a \in A, b \in B}\right\}$

that is, the set of all ordered pairs $\left({a, b}\right)$ such that $a \in A$ and $b \in B$.

Now:
 * Elements of $A \times \left({B \times C}\right)$ are in the form $\left({a, \left({b, c}\right)}\right)$;
 * Elements of $\left({A \times B}\right)\times C$ are in the form $\left({\left({a, b}\right), c}\right)$.

So for $A \times \left({B \times C}\right) = \left({A \times B}\right)\times C$ we would need to have that $a = \left({a, b}\right)$ and $\left({b, c}\right) = c$.

This can not possibly be so, except perhaps in the most degenerate cases.

So from the strict perspective of the interpretation of the pure definitions, $A \times \left({B \times C}\right) \ne \left({A \times B}\right) \times C$.

Comment
Despite this result, the cartesian product of three sets is usually just written $A \times B \times C$ and understood to be the set of all ordered triples.

That is, as the set of all elements like $\left({a, \left({b, c}\right)}\right)$.

From Cardinality of Cartesian Product, we have that:
 * $\left|{A \times \left({B \times C}\right)}\right| \sim \left|{\left({A \times B}\right)\times C}\right|$

and so:
 * $A \times \left({B \times C}\right) \sim \left({A \times B}\right)\times C$

where $\sim$ denotes set equivalence.

So it matters little whether $A \times B \times C$ is defined as being $A \times \left({B \times C}\right)$ or $\left({A \times B}\right)\times C$, and it is rare that one would even need to know.