# Yes, $0!=1$, but... by convention? [closed]

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.

• Take a look at Gamma function May 30, 2021 at 12:09
• Check this: Prove $0! = 1$ from first principles, or these linked questions: math.stackexchange.com/questions/linked/20969 May 30, 2021 at 12:09
• Since $n!=(n-1)! n$, setting $n=1$ shows a consistent value for $0!$
– Sal
May 30, 2021 at 12:10
• Does this answer your question? Prove $0! = 1$ from first principles
– Blue
May 30, 2021 at 12:11
• An empty sum is $0$ and an empty product is $1$, conventionally. Each time, it is the neutral element of the corresponding operation. May 30, 2021 at 12:19

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 .