Why is 0! = 1?

A common question when students first encounter the idea of the factorial, comes from the fact that zero factorial is 1. That is to say, \(0! = 1\).

A quick description of factorial

I will define the factorial in the following way, starting with some examples: 2! is the number of lists of length 2 containing the numbers from 1 to 2. There are just 2 such lists. They are [1,2] and [2,1]. Therefore

$$ 2! = 2$$

The number 3! is the number of lists containing exactly the number 1 to 3. There are six such lists.

[1,2,3]

[1,3,2]

[2,1,3]

[2,3,1]

[3,1,2]

[3,2,1]

Therefore

$$ 3! = 6 $$

In general \(n!\) is the number of lists containing 1 to n. Here n is any positive integer.

One can prove that in general, \(n! = 1\cdot 2\cdot 3\cdots n\). That is to say, you can calculate \(n!\) by multiplying all the numbers up to n.

For example \(2! = 1\cdot 2 = 2\). Also \(3! = 1\cdot 2\cdot 3 = 6\).

This means that the number of lists containing 1 to 4 is

$$ 1\cdot 2\cdot 3\cdot 4 = 24 $$

I won’t try to list them all, but you should feel free to find them all!

0! = 1

Now we can discuss 0!, and actually, the discussion becomes very easy now that we have our rigorous definition.

0! refers to the number of lists with no numbers in them. That is to say, we want to count the number of lists with nothing in them.

There is just one list like that: []. The empty list.

Therefore

$$ \(0!=1\) $$