Zero factorial equals 1 because 0! is the product of no numbers at all — the “empty product” — and an empty product equals 1, the multiplicative identity. The recursion n! = n × (n−1)! forces the same answer: setting n = 1 gives 1! = 1 × 0!, so 0! must be 1.
It feels wrong at first — multiplying by zero always gives zero. But 0! never multiplies by zero: the factorial of n multiplies the numbers from 1 up to n, and for n = 0 that list is empty. Below are three proofs, the mistake that traps students, and one sense of scale: 52! is about 8.066 × 10^67.
What Is a Factorial?
The factorial of a positive whole number n, written n!, is the product of every positive whole number from 1 up to n: n! = 1 × 2 × 3 × … × n. So 5! = 1 × 2 × 3 × 4 × 5 = 120, 3! = 6, 2! = 2, and 1! = 1. Factorials count arrangements: 5! = 120 is the number of ways to order 5 distinct objects, which is why factorials sit at the heart of permutations and combinations.
Apply the definition to 0 and the product contains no numbers at all — an empty product. By long-standing convention, the same one that makes an empty sum equal 0, an empty product equals 1, because 1 changes nothing when you multiply by it. Check any value instantly with a free factorial calculator, which shows the multiplication step by step.
Proof 1: Walk the Pattern Downward
Follow the factorials downward and see where the pattern is forced to land.
Worked example: walk from 5! down to 0!.
- Start with the values you know. 5! = 120, 4! = 24, 3! = 6, 2! = 2, 1! = 1.
- Spot the recursion. Each step down divides by the number you just passed: 120 ÷ 5 = 24, 24 ÷ 4 = 6, 6 ÷ 3 = 2, 2 ÷ 2 = 1. In symbols, n! = n × (n−1)!, so (n−1)! = n! ÷ n.
- Take one more step. At n = 1: 0! = 1! ÷ 1 = 1 ÷ 1 = 1.
- Check the consistency. With 0! = 1, the recursion 1! = 1 × 0! = 1 × 1 = 1 holds. With any other value, the rule breaks at the very first step.
Proof 2: Count the Ways to Arrange Nothing
Factorials count arrangements. The number of ways to arrange n distinct objects in a row is n!: 3 objects have 3! = 6 orders (ABC, ACB, BAC, BCA, CAB, CBA), 2 objects have 2! = 2, and 1 object has 1! = 1 — it just sits there.
How many ways are there to arrange zero objects? There is nothing to move — but there is still exactly one way for the arrangement to be: the empty arrangement, doing nothing. Every counting formula needs the zero-object case to count as one, not zero: choosing all n items from n is obviously possible in exactly one way, and it would come out as zero otherwise.
Proof 3: The Combination Formula Demands It
The number of ways to choose r items from n, written nCr, is nCr = n! ÷ [r! × (n − r)!]. Choose all 3 items from a set of 3: common sense says there is exactly one way — take everything. Plug n = 3, r = 3 into the formula: 3C3 = 3! ÷ [3! × (3 − 3)!] = 6 ÷ [6 × 0!]. For this to equal 1, the denominator must be 6 — so 0! = 1. If 0! were 0, the formula would divide by zero; if it were anything else, 3C3 would come out wrong. The same check works with any numbers: 2C2 and 5C5 = 120 ÷ (120 × 0!) = 1 both force 0! = 1. The definition is not arbitrary — it is the only value that keeps combinatorics consistent.
The parallel with powers is worth noting: x^0 = 1 for the same reason — it keeps the laws of exponents working when the exponent can no longer mean “repeated multiplication.” The exponent calculator shows the same x^0 = 1 behaviour for any base.
How Big Is 52! — And What It Is Not
Factorials grow shockingly fast. 52! ≈ 8.066 × 10^67, a 68-digit number — the number of ways to shuffle a standard deck of 52 cards, so every properly shuffled deck is almost certainly an arrangement no one has ever seen before.
One popular comparison gets it wrong: 52! is not larger than the number of atoms in the observable universe (roughly 10^80); it is about 10^13 times smaller. The true comparison: there have been only about 4.35 × 10^17 seconds since the Big Bang, so 52! is roughly 1.85 × 10^50 times the age of the universe in seconds.
3 Mistakes That Cost Marks
1. Saying 0! = 0 “because zero is in it.” Zero is never a factor of 0! — the factorial multiplies the numbers from 1 up to n, and for n = 0 that list is empty. The confusion comes from mixing up 0! with 0 × something, a different expression entirely.
2. Treating 0! as undefined. Factorials of negative integers are undefined, but 0! is well defined: it equals 1. On an exam, “undefined” loses the mark.
3. Forgetting 0! inside nCr and nPr. Writing 5C5 = 5! ÷ 5! drops the (n − r)! term when r = n. Always write the full formula: 5C5 = 120 ÷ (120 × 1) = 1.
Frequently Asked Questions
Why is 0 factorial equal to 1?
Because 0! is the empty product, which equals 1 by convention, and the recursion n! = n × (n−1)! forces 0! = 1 when n = 1. It is also the only value that keeps the combination formula working: 3C3 must equal 1, which requires 0! = 1.
Is 0! equal to 0 or 1?
1. The value 0 would come from multiplying by zero, but 0! never multiplies by zero — it multiplies an empty list of numbers, and the empty product is defined as 1.
What is the proof that 0! = 1?
Walk the pattern down: 5! = 120, 4! = 24, 3! = 6, 2! = 2, 1! = 1, each step dividing by n. One more step gives 0! = 1! ÷ 1 = 1. The combination formula agrees: 3C3 = 6 ÷ (6 × 0!) must equal 1, so 0! = 1.
Why is 0! not undefined like division by zero?
Because no division by zero is involved — 0! is defined directly as the empty product, and every factorial formula stays consistent with it.
How do you calculate 0! on a calculator?
Enter 0 and press the factorial key (often labelled x! or n!) — a correct calculator returns 1. The question “why is 0 factorial 1” is exactly what Calcrio’s factorial calculator answers with the worked steps shown.
Keep Practicing
Work these by hand, then verify each with the free factorial calculator: 4! = 24; 6! ÷ 3! = 720 ÷ 6 = 120; 10C10 = 10! ÷ (10! × 0!) = 1; and 0! × 5! = 1 × 120 = 120.