Why is $0! = 1$ the same as $1! = 1$?

Question

I want to ask why is $$0! = 1$$ the same as $$1! = 1.$$

As a student I was lost and when I tried to ask the question the teacher said this will be done in complex analysis.

I know here I will thirst my quest by the help of people who understand it.

I know We define $n!$ as the product of all integers k with $1\leq k\leq n$. When $n=0$ this product is empty so it should be 1.

I was just wondering if for factorial we take:

$$6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$$ $$5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$$

In normal multiplication any number multiplied by $0$ = $0$

hence $$0! = 0\cdot 0 = 0$$ hence $$1! = 1\cdot 1 = 1$$

I am confused by this help Thanks.

But then the factorial rule of $0!$ being $1$ is misleading because $0$ x emptyset = emptyset why would it be 1?

Answer 1

The number $n!$ counts the number of bijections from a set of $n$ elements to itself. There is exactly one bijection from the emptyset to itself, the empty function. Thus $0!=1$.

Answer 2

You should watch this Numerphile video on Youtube:

https://www.youtube.com/watch?v=Mfk_L4Nx2ZI

Here the argument is made that $0!$ should be one because then the rule $$ n! = \frac{(n+1)!}{n+1} $$ is true even for $n=0$. One can make other arguments, this is just one.

Answer 3

Note that with the usual definition, $n!=n(n-1)(n-2)\cdots1$, we have $n!=n(n-1)!$

If we extend this using $n=1$, we get $1!=1\cdot0!$.

It is also useful to define a product of $0$ numbers to be $1$. This also works for $0!$ which would be a product of $0$ integers.

The Binomial Theorem also works when we define $0!=1$, for example $$ \binom{n}{0}=\frac{n!}{(n-0)!\,0!}=1 $$

Answer 4

You have written $$6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\ 5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$$ and so on, but then concluded $$0! = 0 \cdot 0.$$ This doesn't fit the pattern, which should continue as follows: $$4! = 4\cdot 3\cdot 2\cdot 1\\ 3! = 3\cdot 2\cdot 1\\ 2! = 2\cdot 1\\ 1! = 1\\ 0! = $$

At this point you simply need to define what $0!$ should mean, and it is natural to choose $1$, the multiplicative identity.

Answer 5

We have the gamma function that is defined for all complex number $z$, excluding the non-positive integers, specifically $$\Gamma(z) = \int_0^{\infty}t^{z-1}e^{-t} \, \mathrm{d}t$$

We can then define the factorial as $$n! = \Gamma(n+1).$$ For $n=0$, we have $$0! = \Gamma(1) = \int_0^{\infty} e^{-t} \, \mathrm{d}t = 1$$

Which is an improper integral that converges to $1$. So we have $0! = 1$.

---

It is quite easy to see that the integral converges to $1$, we need only simply evaluate the definite integral as follows $$\Gamma(1) = \lim_{a \to \infty} \int_0^{a} e^{-t} \, \mathrm{d}t = \lim_{a \to \infty}\left[-e^{-t}\right]_0^a = \lim_{a\to \infty} \left(1 - e^{-a}\right) = 1.$$

Answer 6

We have by definition $(n+1)! = (n+1)\,n!$. Therefore $1! = (0+1)! = (0+1)\,0! = 0!$

Answer 7

You can think of the definition of the factorial backwards: $$(n-1)!=\frac{n!}{n}.$$ Then you get \begin{align} 2!&=\frac{1\cdot 2\cdot 3}{3}=1\cdot 2,\\ 1!&=\frac{1\cdot 2}{2}=1,\\ 0!&=\frac{1}{1}=1. \end{align} Voilà!

Answer 8

A power of exponent $0$ and positive base is $1$. This means that a product of no factors is $1$