Factorial Notation

You have 5 books on a shelf — a maths textbook, a physics textbook, a chemistry textbook, a novel, and a dictionary. You want to rearrange them. In how many different orders can the 5 books stand?

For the first position, you have 5 choices. Once you place a book there, 4 remain for the second position. Then 3, then 2, then 1. By the multiplication principle:

That product — 5 \times 4 \times 3 \times 2 \times 1 — appears so often in mathematics that it has its own notation: 5!, read "five factorial." Every time you arrange n distinct objects in a row, the answer is n!. Every time you count the ways to fill n slots where the choices shrink by one at each step, n! appears. The exclamation mark is not hyperbole — it is earned. Factorials grow faster than exponentials, faster than polynomials, faster than almost anything you have seen so far.

Definition of n!

The first few values:

By 10! you already have over three million — roughly the population of Jaipur. By 15! the value exceeds one trillion. By 20! it is larger than the number of grains of sand on Earth (estimated at about 10^{18}), since 20! = 2{,}432{,}902{,}008{,}176{,}640{,}000 \approx 2.4 \times 10^{18}.

Why 0! = 1. This is not a trick and not arbitrary. There is exactly one way to arrange zero objects: do nothing. The empty arrangement exists — it is the unique sequence of length 0. Defining 0! = 1 also makes every factorial formula consistent. For instance, the recursion (n+1)! = (n+1) \times n! gives 1! = 1 \times 0!, so 0! must be 1 for the formula to hold at n = 0. The number of subsets of a set with n elements is 2^n, and the formula for choosing 0 elements from n is \binom{n}{0} = \frac{n!}{0! \cdot n!}, which equals 1 only when 0! = 1. Every algebraic and combinatorial road leads to the same answer.

Properties of the factorial

Recursive property. The most useful single fact about factorials:

This follows directly from the definition: (n+1)! = (n+1) \times n \times (n-1) \times \dots \times 1 = (n+1) \times n!. It works for all n \geq 0:

• 1! = 1 \times 0! = 1 \times 1 = 1
• 2! = 2 \times 1! = 2 \times 1 = 2
• 3! = 3 \times 2! = 3 \times 2 = 6
• 4! = 4 \times 3! = 4 \times 6 = 24

The recursion is how you compute factorials step by step, and it is how you simplify factorial expressions algebraically.

Cancellation. The recursion lets you cancel common factorial factors in fractions. For instance:

More generally:

This is a product of r consecutive integers descending from n. It appears constantly in permutations.

Divisibility. n! is divisible by every integer from 1 to n, because each of those integers appears as a factor in the product. In fact, n! is divisible by the product of any k consecutive integers from 1 to n.

The recursion (n+1)! = (n+1) \times n!

The recursion deserves its own section because it is the single most-used tool for working with factorials.

Simplifying ratios. Express \frac{(n+2)!}{n!} in terms of n:

So \frac{(n+2)!}{n!} = (n+2)(n+1).

Solving equations. If \frac{n!}{(n-2)!} = 42, what is n?

So n(n-1) = 42. This gives n^2 - n - 42 = 0, which factors as (n-7)(n+6) = 0. Since n must be a positive integer, n = 7.

Check: \frac{7!}{5!} = 7 \times 6 = 42. Correct.

Building the staircase. Every value of n! is built from the previous one by a single multiplication. Think of it as a staircase: to go from n! to (n+1)!, you multiply by (n+1). To go backwards — from (n+1)! to n! — you divide by (n+1). The recursion is the single step of the staircase, and the full factorial is the result of climbing from 0! = 1 all the way up to n.

Exponent of a prime in n!

Here is a question that connects factorials to number theory: what is the highest power of a prime p that divides n!?

Take 10! and the prime 2. The product 10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 contains several factors of 2:

• Multiples of 2 in \{1, \dots, 10\}: 2, 4, 6, 8, 10 — that is \lfloor 10/2 \rfloor = 5 numbers, contributing at least one factor of 2 each.
• Multiples of 4 = 2^2: 4, 8 — that is \lfloor 10/4 \rfloor = 2 numbers, contributing an extra factor of 2 each.
• Multiples of 8 = 2^3: 8 — that is \lfloor 10/8 \rfloor = 1 number, contributing yet another extra factor of 2.
• Multiples of 16 = 2^4: none in \{1, \dots, 10\}, so \lfloor 10/16 \rfloor = 0.

Total exponent of 2 in 10!: 5 + 2 + 1 + 0 = 8.

Check: 10! = 3{,}628{,}800 = 2^8 \times 3^4 \times 5^2 \times 7. The exponent of 2 is indeed 8.

This method works for any prime p and any n, and it is called Legendre's formula.

Why it works. Each term \lfloor n/p^k \rfloor counts how many integers from 1 to n are divisible by p^k. Each such integer contributes at least k factors of p to the product n!. But counting "multiples of p" already counted one factor from each; counting "multiples of p^2" adds the second factor, and so on. The sum adds up all the contributions without double-counting.

Two worked examples

Common confusions

• "0! should be 0 because there is nothing to multiply." The product of no numbers is 1 (the empty product), not 0. This is the same convention that makes x^0 = 1 in exponents and powers — the empty product is the multiplicative identity.

• "n! means n multiplied by itself." No — n! is the product of all integers from 1 to n, not n \times n = n^2. So 4! = 4 \times 3 \times 2 \times 1 = 24, not 4 \times 4 = 16.

• "(a + b)! = a! + b!." This is almost never true. 3! + 2! = 6 + 2 = 8, but (3 + 2)! = 5! = 120. The factorial does not distribute over addition.

• "(a \cdot b)! = a! \cdot b!." Also false in general. (2 \cdot 3)! = 6! = 720, but 2! \cdot 3! = 2 \times 6 = 12.

• "The number of trailing zeros in n! is \lfloor n/10 \rfloor." Close but wrong. The correct answer uses Legendre's formula with p = 5: you need \lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \dots, because multiples of 25, 125, etc. contribute extra factors of 5. For n = 100, \lfloor 100/10 \rfloor = 10 but the correct answer is 24.

• "Factorials are only defined for positive integers." Standard factorials are defined for non-negative integers (so 0! is included). There is an extension to non-integer and even complex inputs called the Gamma function, where \Gamma(n+1) = n! for non-negative integers, but that is beyond the scope of this article.

Going deeper

Stirling's approximation

For large n, computing n! exactly is impractical — the numbers are enormous. But an excellent approximation exists:

This is Stirling's formula. For n = 10: \sqrt{20\pi} \cdot (10/e)^{10} \approx 7.926 \times 3{,}628{,}800 / 3{,}628{,}800 \approx 3{,}598{,}696, while 10! = 3{,}628{,}800 — the approximation is off by about 0.8\%. The accuracy improves as n grows. Stirling's formula shows that n! grows roughly like (n/e)^n, which is faster than a^n for any fixed base a — factorial growth dominates exponential growth.

Proof by induction that n! = n \times (n-1)!

The recursion is so obvious from the definition that it barely needs a proof, but writing one anyway is good practice with mathematical induction.

Base case: n = 1. The claim is 1! = 1 \times 0!. Indeed, 1! = 1 and 1 \times 0! = 1 \times 1 = 1.

Inductive step: Assume n! = n \times (n-1)! holds for some n \geq 1. Then:

The step follows directly from the definition of (n+1)! as the product of all integers from 1 to n+1. The inductive hypothesis was used to identify the bracketed portion as n!.

Wilson's theorem: a connection to primes

Here is a surprising link between factorials and primes. An integer p > 1 is prime if and only if:

For example, 4! = 24 \equiv -1 \pmod{5} (since 24 = 5 \times 5 - 1), confirming that 5 is prime. And 5! = 120 \equiv 0 \pmod{6} (since 120 = 6 \times 20), confirming that 6 is not prime. Wilson's theorem is beautiful but computationally useless for testing primality, because computing (p-1)! is harder than simply checking divisibility. Its value is theoretical — it shows that the factorial carries deep structural information about primes.

Where this leads next

The factorial is a building block for the entire combinatorics sequence.

• Permutations — Basics — the formula {}^nP_r = \frac{n!}{(n-r)!} is a direct application of the factorial and the multiplication principle.
• Fundamental Principle of Counting — the article that built the multiplication principle, which is the reason n! counts arrangements.
• Exponents and Powers — where the law a^0 = 1 mirrors the convention 0! = 1, both as empty products.
• Number Theory Basics — Legendre's formula connects the factorial to prime factorisation, and Wilson's theorem links it to primality.
• Mathematical Induction — the technique for proving factorial identities rigorously, including the recursion itself.