Prove $0! = 1$ from first principles

Question

How can I prove from first principles that $0!$ is equal to $1$?

Answer 1

I'm not sure that there is anything to prove. I think it follows directly from the definition of factorial:

$$ n! := \prod_{k = 1}^n k$$

So if $n=0$ the right hand side is the empty product which is $1$ by convention.

Answer 2

We need $0!$ to be defined as $1$ so that many mathematical formulae work. For example we would like $$n! = n \times (n-1)!$$ to work when $n=1,$ ie $1! = 1 \times 0!.$ Also we require that the formula for the number of ways of choosing $k$ objects from $n$ is valid for $k=n.$ ie $${n \choose k} = \frac{n!}{k!(n-k)!}$$ is valid when $k=n.$

Things need to work when we extend our definition of the factorial via the gamma function.

$$\Gamma(z) = \int\limits_0^\infty t^{z-1} e^{-t} \,\mathrm{d}t,\qquad \Re(z)>0.$$

The above gives $\Gamma(n)=(n-1)!$ and so we require $0!=1,$ since $\Gamma(1)=1.$

Answer 3

One of the simplest ways of doing this is to observe that if you have $$ 6!= 720 $$ then divide both sides by 6 to get $$ 5!=120 $$ then divide both sides by 5 to get $$ 4!=24 $$ then divide both sides by 4 to get $$ 3!=6 $$ then divide both sides by 3 to get $$ 2!=2 $$ then divide both sides by 2 to get $$ 1!=1 $$ then divide both sides by 1 to get $$ \text{[fill in the blank here]} $$

Answer 4

The empty product is taken to be equal to 1. Take logs and you get an empty sum equal to zero, which is somehow more intuitive, but this trick of taking logs to convert a product into a sum never seems to get a mention in the literature. [Assumes products have positive terms]

Answer 5

One can prove this by convention (that is based on what 'works', based on base cases for product of a list).

But for a proof from 'nothing' as it were...define what factorial is supposed to mean. I think of it as the number of one-to-one, onto (=bijective) functions from a set of size $n$ to itself. if$n$ is 0 then the set is the empty set. How many bijective functions are there from $\emptyset$ to $\emptyset$? (a function is a set of ordered pairs (with some restrictions)? There are no legal pairs for these restrictions, and so there are no legal subsets of these pairs...other than the empty set. So $\emptyset$ -is- a function (it -is- a set of pairs (empty of course), all of whose pairs satisfy the function criteria). No other functions will work so $\emptyset$ is the only function that works. So there is only 1 bijective function on a set of size 0. So $0! = 1$.

Yes, this is weird, but it works. Negative numbers, complex numbers, they're all weird even when you just manipulate their properties. But you'll get over it.

Answer 6

$0! = 1$ is consistent with, and for reasons related to, how we define the empty product.
See, for example, this entry on empty product. This is simply the name of the phenomenon Michael Hardy alludes to:

Empty product:

The empty product of numbers is the borderline case of product, where the number of factors is zero, that is, the set of the factors, is empty. In such a "borderline" case, the empty product of numbers is equal to the multiplicative identity, which is $1$.

Some of the most common examples are the following:

• The zero$^{\text{th}}$ power of a number $a$: $a^0 = 1$,
• The factorial of $0$: $0! = 1$,
• The prime factor presentation of unity, which has no prime factors.

Just as ${n^0 = 1}$ for any $n$, and the "prime factorization" of $1$ = $1$, we define, as a matter of convention, $0! = 1$.

Answer 7

Let's try a different approach from my other answer to this:

To multiply a number $N$ by $6!$ is to multiply it by six factors and get $$ N\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6. $$ Similarly to multiply $N$ by $0!$ is to multiply it by $0$ numbers: $$ N. $$ But that is the same as multiplying it by $1$ $$ N\cdot1. $$ Multiplying $N$ by no numbers at all is multiplying $N$ by $1$.

(This answer doesn't apply only to factorials; it may be taken as a general explanation of why, when one multiplies no numbers, one gets $1$.)

Answer 8

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

Explanation 2: If $n$ is a nonnegative integer, we define $n!$ to be the number of orderings on a set with $n$ distinct objects. If $n = 0$, this set is empty. Vacuously, it has 1 order.

Answer 9

You can define $\exp(x)$ as

$$ 1 + \frac {x} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... $$ but the following seems more uniform: $$ \frac {x^0} {0!} + \frac {x^1} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... $$ These are only equal if $0! = 1$

Not a proof, but makes sense to me!