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I have seen many math books, and some of them, very good books, that say that $0!=1$ 'by convention'. I think that $0!$ must be $1$ because it is the product of the empty set. That is, for $a\neq 0$, $$a^0=0!=\prod\emptyset=1$$ What do you think?

EDIT: I'm not asking for a proof that $0!=1$, I really don't be needed to prove that. The question is, again, 'what do you think?', that is, do you think that is more convenient to define $0!=1$ by... well, convention, or is it better to see $0!$ as an empty product?

In other words, this is a soft question.

ajotatxe
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2 Answers2

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One way to define $n!$ is that it’s the number of bijections from a set with $n$ elements to itself ,If the set is empty then the number of elements is $0$ hence $0!=1$ ,since there is only one bijection from the empty set to itself .
It is also notationally useful for example the power series of $\exp(x)$ is $\sum_{0}^{\infty}x^n/n!$ this makes use of $0!=1$ .
Another advantage is in combinatorics the number of ways to arrange $n$ objects is $n!$ , so the number of ways to arrange $0$ objects is $1$ ,this makes sense .

Vivaan Daga
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I think if you are teaching factorial in a procedural way then trying to.explain the product of an empty set is pretty far outside the context, and simply saying it is a matter of convention is more cognitively in line with where the student is at.

When teaching at the board, I try to separate out what a factorial is (number of ways a set can be rearranged) vs how to calculate it. Asking the class to tell me how many different ways they can arrange an empty set usually gives them the picture.

johnnyb
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