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In many places, I have read that $0!$ is $1$ but as I know that factorial is multiple of the number and all the number before it up to $1$. But according to it $0!$ should be $0\times0$ that is $0$ then why it is $1$ . Please explain in detail.

Cheese Cake
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Jatin
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    You can't prove a definition. You can justify why you are adopting it. – Sassatelli Giulio Sep 05 '22 at 08:17
  • $n!$ is the product of the integers $k$ such that $1\le k\le n$. I don't see why the product of the integers $k$ such that $1\le k\le 0$ should be $0\times 0$. – Sassatelli Giulio Sep 05 '22 at 08:19
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    By the previous comment, $0!$ is the empty product. – Anne Bauval Sep 05 '22 at 08:23
  • for n > 1, $n! = (n-1)!$, so by convention 0! is considerd to be 1. And there are i suppose others justifications for exmple because it is usufull for writting some formula implying factorial – MrSmithGoesToWashington Sep 05 '22 at 08:24
  • If we would include $0$ as well, EVERY factorial would be $0$. So , of course , $0$ is omitted. And then there are no positive integers upto $0$ and the empty product is , per definition , $1$ , hence $0!=1$ is justified. – Peter Sep 05 '22 at 10:32

3 Answers3

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$$\frac{n!}{(n-1)!} = n.$$ Set $n=1$ and solve for the denominator.

Mauro ALLEGRANZA
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David G. Stork
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Take a recursive definition of integer factorial as

$$n!=\begin{cases} 1,& \text{ if } n=1\\ n\cdot(n-1)!,& \text{ if } n>1\\ \end{cases}\tag 1$$

so that $1!=1$, $2!=2$, $3!=6$ etc.

Now you can use the defintion in "reverse":

$$(n-1)! = \frac{n!}n\tag 2$$

Which yields $$0!=(1-1)!= \frac{1!}1 = 1\tag 3$$

However, you cannot go any further, because for $(-1)!$ yuo would divide by $0$ in $(2)$. This reflects the fact that the Gamma function has a pole at $0$. Remember that $\Gamma(n) = (n-1)!$.

As an aside, $(3)$ means that we could have defined factorial starting at $n=0$ like

$$n!=\begin{cases} 1,& \text{ if } n=0\\ n\cdot(n-1)!,& \text{ if } n>0\\ \end{cases}\tag 4$$

which yields the exact same values like $(1)$ plus one backstep.

emacs drives me nuts
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There's no proof for why $0! = 1$ because that's a definition, but we can at least explain why we've adopted it. Here are some reasons why.

(1) - There is exactly one permutation of zero objects: with nothing to permute, the only "rearrangement" is to do nothing.

(2) - In calculus, this definition matches the gamma function ${\displaystyle 0!=\Gamma (0+1)=1}$, and the gamma function must have this value to be continuous.

(3) - In calculus, this definition is convenient for power series representations like $e^x = \displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}$

(4) - If $n = 0$, then the definition of $n!$ as a product of nonnegative integers involves the product of no nonnegative integers at all. In fact, the empty product, which is a product of no factors, is equal to the multiplicative identity.

Accelerator
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