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How to define (rigorously) ordered tuples and finite cartesian product in ZFC ? Does it possible to extend the Kuratowski definition of ordered pair, defining as $(x,y) = \{\{x\},\{x,y\}\}$... However, for pair, the Schema of comprehension ensure us the existence and uniquness (extensionality) of the cartesian product of $A$ and $B$, which can be defined as $A\times B = \{ (x,y)\in\mathcal P(\mathcal P(A\cup B)) : x\in A \land y\in B\}$... but for $n$-tuple, how does it work?

I want a formal definition of "what a $n$-tuple is", in order to define the cartesian product $E_1\times \cdots \times E_n$ using the Axiom Schema of Comprehension (as $E_1\times \cdots \times E_n := \{ (x_1, \ldots, x_n) \in \,?\,: x_1\in E_1 \land \cdots \land x_n \in E_n\} $ where the "?" set is a certain power set and depends on $n$-tuple definition).

Ideally, without any recursive definition like $(x_1, \ldots, x_{n-1}, x_n) = ((x_1, \ldots, x_{n-1}),x_n)$...

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    $\{\{x\},\{x,y\},\{x,y,z\},\{x,y,z,w\}, \dotsc\}$? – 2017-02-26
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    Asking for non-recursive definitions in set theory seems a bit counter to the whole idea, though... – 2017-02-26
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    I thought about it, but I'm unable to "formalize" it with $\bigcup$... :/ – 2017-02-26
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    @Chappers: $(x,x,y)$ and $(x,y,y)$ and $(x,y,x)$ will all give you $\{\{x\},\{x,y\}\}$. So the "obvious generalization" obviously fails. – 2017-02-27
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    Relevant: http://math.stackexchange.com/questions/25791/definition-of-an-ordered-pair and http://math.stackexchange.com/questions/307168/about-the-definition-of-n-tuple – 2017-02-27

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Writing $\langle a,b \rangle$ for the ordered pair, we may define $$\langle x_1, \dots, x_n \rangle := \{ \langle m, x_m \rangle : 1 \leq m \leq n \}$$ though this is not standard; I believe the standard way to do it is recursively, because that's very natural and doesn't require these privileged "natural number" objects.