Proof of $0! = 1$

Question

I have been recently studying binomial theorem and in that we very frequently encounter factorials. But one queer thing which I found is $0!$. Even more queer is its value which is

$0! = 1$.

I was told this fact and I accepted it(without any proof). But I am still confused about it.

How can we prove that $0!= 1 $?

It's the definition of the factorial at $n=0$, so there's nothing to prove. — ireallydonknow, Jan 01 '14 at 17:55
Basically, this is just a convention. But notice how nicely this agrees with general formula $n!=n\cdot(n-1)!$ (take $n=1$). — Grigory M, Jan 01 '14 at 17:57
@ireallydonknow So is it really by definition? Is there no proof to it? — user2369284, Jan 01 '14 at 17:57
@user2369284 There isn't a proof as this is the definition. It's a definition we make so that we can make formulas a lot simpler for ourselves. Think about this: why do we say that $1$ is not prime? On the surface, it doesn't seem like we need to, but we say that $1$ isn't prime because it makes theorem statements simpler. Similar reasoning applies here: we don't need to but it makes statements simpler. — Cameron Williams, Jan 01 '14 at 18:01
For any cardinal $c$ the notation $c!$ designates the order of the group of permutations of a set of cardinal $c$. — Pierre-Yves Gaillard, Jan 01 '14 at 18:08
What is your definition of $n!$? Without a definition you cannot prove anything. — Carsten S, Jan 01 '14 at 18:16
As the local worshipper of simplicity, I hope my answer wins the "Keep It Simple" award. — Michael Hardy, Jan 01 '14 at 18:23

Antoine · Accepted Answer · 2014-01-01T18:37:46.170

7

We can order $n$ elements in $n!$ ways. $0$ elements can be ordered in just one way. One can choose $k$ of $n$ elements in ${n \choose k} = \frac{n!}{k!(n-k)!}$ ways. Obviously, $$1 = {n\choose n} = \frac{n!}{n!\cdot 0!}\text{.}$$ Also, $0! = 1$ agrees with Euler's function $\Gamma$, for which $\Gamma(n) = (n-1)!$ holds.

These are not proves, but only good reasons why should $0!$ equal $1$.

edited Jan 01 '14 at 18:37

answered Jan 01 '14 at 18:03

Antoine

3,439

You can also include the point(as others have posted) that $0! = 1$ by definition as $n! = \frac{(n+1)!}{n+1}$. – user2369284 Jan 01 '14 at 18:18

benh · Answer 2 · 2014-01-01T18:14:39.253

As the other answers pointed out, $0! = 1$ is rather a definition than a thing we could prove, but there is some motivation behind it:

First of all, it is a common convention, that the empty sum $\sum_\emptyset=0$ has value $0$. It seems to be a natural definition, because if we add something to the sum, we want to be able to sum by forming the disjoint union of the two sets: $$\sum_{x\in A} x + \sum_{x \in \emptyset} x = \sum_{x \in A \cup \emptyset} x = \sum_{x \in A}.$$ In the same way, we often define $\bigcup_\emptyset =\emptyset$ and $\prod_\emptyset = 1$. The latter gives us: $$n!:=\prod_{k=1}^nk \Rightarrow 0! = \prod_{k=1}^0 = \prod_\emptyset k = 1.$$ Thus, setting $0!$ is consistent to other conventions.
Another answer comes from real analysis. By the Bohr-Mollerup Theorem, there is only one function $f:\Bbb R^+ \rightarrow \Bbb R$ that satisfies:
- $f(x+1)=xf(x) \forall x \in \Bbb R$
- $f$ is logarithmic convex
- $f$ interpolates the factorial in a sense that for $n \in \Bbb N$: $f(n+1) = n!$
This function is the Gamma-function and it has value $\Gamma(1) = 1$ corresponding to $0! = 1$. Thus setting $0! = 1$ can also be considered to be a natural choice.

score 2 · Answer 3 · answered Jan 01 '14 at 18:01

You can prove it from the definition of the factorial operation, which defines $0!=1$. The question you should be asking is why we define it this way.

One argument in favor of this definition is that it preserves the important property that $$(n+1)! = (n+1)\cdot n!$$ even when $n=0$. If we want to use this property to reason about factorials, we either need to accept that $0!=1$ or else be prepared to fill our arguments with special exceptions for $n=0$.

A different kind of argument proceeds from the objects that the factorial function is intended to model. $n!$ is intended to be the number of ways of making an ordered list of $n$ distinct objects. So for example $3!=6$ because there are six ways to order a list of three items:

$$\begin{array}{ccc} (123) & (132) & (312) \\ (213) & (231) & (321) \end{array}$$

And $2! = 2$ because there are two ways to order a list of two items:

$$\begin{array}{cc} (12) & (21)\end{array}$$

And $1!=1$ because there is one way to order a list of one item:

$$\begin{array}{c} (1)\end{array}$$

And $0!=1$ because there is one way to order a list of zero items:

$$\begin{array}{c} ()\end{array}$$

$()$ is unquestionably a list of zero items, so why wouldn't you count it?

score 1 · Answer 4 · answered Jan 01 '14 at 18:00

1

Note that:

$$n! = \frac{(n+1)!}{n+1}$$

So you can apply this for $0$. Actually it's quite nicely explained here

answered Jan 01 '14 at 18:00

Stefan4024

35,843

Michael Hardy · Answer 5 · 2014-01-01T20:24:35.173

1

\begin{align} 4! & = 24. \quad \text{Divide by 4 and get:} \\ 3! & = 6. \quad \text{Divide by 3 and get:} \\ 2! & = 2. \quad \text{Divide by 2 and get:} \\ 1! & = 1. \quad \text{Divide by 1 and get:} \\ 0! & = \end{align}

Fill in the blank.

Not a proof, but this may shed some light.

Also, notice that when you multiply no numbers at all, you get $1$. That's also been asked about here. That's why $x^0=1$.

edited Jan 01 '14 at 20:24

answered Jan 01 '14 at 18:22

Michael Hardy

1

score 0 · Answer 6 · answered Jan 01 '14 at 18:05

There is no proof for this statement, since $0! = 1$ is just a convenient and nice definition in many applications.

For instance, if you would like to think of $n!$ as the number of bijective maps from an $n$ element set to itself, then there is only one bijective map from an empty set to itself.

The other to interpret factorial is through the $\Gamma$ function defined as $\Gamma(z+1) = z \Gamma(z)$. If $z \in \mathbb{Z}^+ - \{1\}$, then $\Gamma(z) = (z-1)!$. Hence, if we would like to extend the notion of factorial beyond positive integers using the $\Gamma$ function, setting $z=1$, we get that $1! = 1 \cdot 0!$, which forces $0! = 1$.

Proof of $0! = 1$

6 Answers6