Is Cartesian Product associative?

Let $\displaystyle A, B, C$ be sets.

Is it the case that $\displaystyle A \times (B \times C) = (A \times B) \times C$

where $\displaystyle (A \times B)$ denotes Cartesian Product?

That is, $\displaystyle A \times B = \{(a, b): a \in A, b \in B\}$.

I have J.A. Green's "Sets and Groups" (1965, Routledge & Kegan Paul) saying:

"Show that the sets $\displaystyle A \times (B \times C)$ and $\displaystyle (A \times B} \times C$ are never the same" (Chapter One Exercise 8), on the one hand.

On the other hand, I have W.E. Deskins: Abstract Algebra (1964, Dover) saying:

"Theorem 1.9. $\displaystyle (a, b, c) = (d, e, f)$ if and only if $\displaystyle a=d, b=e, c=f$."

The definition of $\displaystyle (a, b, c)$ in Deskins is given as $\displaystyle (a, (b, c))$, i.e. as a construction of ordered pairs.

This latter sort of seems to imply that $\displaystyle A \times (B \times C) = (A \times B) \times C$, but no mention of it is made in that book.

T.S. Blyth's "Set Theory and Abstract Algebra" (Longman, 1975) gives:

$\displaystyle E \times F \times G = \{(x, y, z): x \in E, y \in F, z \in G\}$

which sort of seems to imply that in this context associativity **does** hold.

I've seen this set as an exercise in several places, but never have I seen a definitive proof one way or another. It hinges on whether $\displaystyle (a, (b, c)) = (a, b, c) = ((a, b), c)$ which you instinctively feel ought to be true, but I can't come up with a convincing argument either way, unless you take *literally* the expression of $\displaystyle (a, b)$ as equal to $\displaystyle \{\{a\}, \{a, b\}\}$ (according to Wiener and Kuratowski) which is in the final instance a convenient way of obtaining an ordered pair in axiomatic set theory (e.g. ZF). And Green's "Sets and Groups" (mentioned above) is pretty assertive when it comes to exercise 1.9, and no mention has been made in that work of the Wiener/Kuratowski definition.

What's the current thinking on this result?