How can I prove from first principles that $0!$ is equal to $1$?
2026-02-24 03:25:58.1771903558
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Prove $0! = 1$ from first principles
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There are 11 best solutions below
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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 $\Gamma(1)=1.$
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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.
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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 $$ \text{[fill in the blank here]} $$
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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]
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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 $1$.)
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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.
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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!
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$0! = 1$ 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, $0! = 1$.
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Perhaps it helps to consider this question in a broader context, where we consider "empty products" more generally. Let me introduce "Capital pi" notation, which is analogous to sigma notation for sums. If $a_1,a_2,\dots,a_{n-1},a_n$ are a list of numbers, then we define $$ \prod_{i=1}^{n}a_i=a_1\cdot a_2\cdot\dots\cdot a_{n-1}\cdot a_n \tag{*}\label{*}\, . $$ The product on the right-hand side of $\eqref{*}$ is obtained by letting $i=1,2,\dots,n-1,n$. A few remarks about this notation will come in handy:
- The letter $i$ is called a "dummy variable", meaning that it is an arbitrary symbol that can be replaced with another letter without changing the value of the product, i.e. $\prod_{i=1}^{n}a_i=\prod_{r=1}^{n}a_r=\prod_{k=1}^{n}a_k$.
- The right-hand side of $\eqref{*}$ should not be understood literally as a product of at least four factors. For instance, $\prod_{i=1}^{3}a_i$ should be taken to mean $a_1\cdot a_2\cdot a_3$.
To make this product notation precise, we require a recursive definition: first we let $\prod_{i=1}^{1}a_i=a_1$, and second we let $\prod_{i=1}^{n}a_i=\left(\prod_{i=1}^{n-1}a_i\right)\cdot a_n$ when $n$ is an integer greater than $1$. Notice that in order to make this definition work, we need to agree that the "product" of a single term $a_1$ to be $a_1$. Therefore, while the "product" of a single term seems like an unnatural idea at first, it is convenient to agree that the product of a single number is itself.
It is for a similar reason that we agree that the product of no numbers at all—the empty product—is equal to $1$. It's not because empty products have to be defined in this way, but rather that this is a convenient notational convention. Why is it convenient? Well, notice that if $n>1$, then $\prod_{i=1}^{n}a_i=\left(\prod_{i=1}^{n-1}a_i\right)\cdot a_n$. If we want this equation to be true when $n=1$, then it ought to be the case that $\prod_{i=1}^{1}a_i=\prod_{i=1}^{0}a_i\cdot a_1$, and so assuming that $a_1$ is non-zero, we must define $\prod_{i=1}^{0}a_i$ as $1$. Of course, we don't want one rule for when $a_1=0$, and one rule for when $a_1\neq 0$, and so the simplest convention to adopt is that $\prod_{i=1}^{0}a_i=1$ regardless of whether $a_1=0$.
How does this relate to factorials? Well, by definition $n!=\prod_{i=1}^{n}i$, and so $0!=\prod_{i=1}^{0}i=1$. In this case, we are motivated by the desire for the equation $n! = n(n-1)!$ to still hold when $n=1$. It's not that $0!$ has to be defined in this way, but rather that this is the most natural definition. This is why it doesn't make much sense to "prove" that $0!=1$; it's not a a matter of proof—it's a matter of definitions.
It turns out that there are a diversity of other reasons why it is convenient to define $0!$ as $1$, which you can read about here. It is fortunate that in every context where $0!$ crops up, it seems obvious what it "should" be. In other cases, we are not so lucky. For instance, sometimes it is convenient to define the degree of the zero polynomial as $-1$, but other times it convenient to define it as $-\infty$, and other times still it feels more natural to leave it undefined. This example shows that mathematical definitions are quite flexible, and that they suit our needs depending on the context, rather than being eternal truths etched into stone.
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.