Zero factorial equals 1, and it is not an arbitrary convention. Four separate arguments force it. The factorial rule n! = n × (n − 1)! gives 1! = 1 × 0!, and since 1! = 1, 0! must be 1. Counting agrees: n! is the number of ways to line up n objects, and there is exactly one way to line up nothing, the empty list. The binomial formula needs it, because there is exactly one way to choose no items from a set. And the gamma function, the smooth curve through the factorials, passes through exactly 1 at n = 0. Every road leads to the same 1.
That still feels suspicious to most people. Zero of anything usually gives zero. The suspicion is worth taking seriously, so here is each argument in turn, then the question of what a “convention” really is.
What a factorial counts
Take n different objects and put them in a row. The number of different orders is n factorial, written n!. We are counting finished rows, not the moves used to make them.
With three objects, A, B and C, there are six orders: ABC, ACB, BAC, BCA, CAB and CBA. You have 3 choices for the first position, 2 for the second, and just 1 left for the third, so 3! = 3 × 2 × 1 = 6. With two objects you get AB or BA, so 2! = 2. With one object there is one arrangement: leave it where it is. Slightly boring, but still an arrangement, so 1! = 1.
Count down and you get 6, 2, 1. The next question is the whole puzzle: with nothing left to arrange, what are we counting?
The exclamation mark itself has a history. The French mathematician Christian Kramp introduced it in 1808, in the preface to his Éléments d’arithmétique universelle, explaining that he used “the very simple notation n!” for the product of the numbers from n down to 1. Not everyone was pleased. In 1842 Augustus De Morgan called it, among other things, a symbol that gives pages “the appearance of expressing surprise and admiration” that 2, 3, 4 and so on should turn up in mathematical results. The notation stayed.
Why is 0! = 1? Run the recursion backwards
There is a pattern hiding inside the count. To arrange n objects, choose which one goes first, in n ways, then arrange the remaining n − 1 objects. That gives the rule
n! = n × (n − 1)!
Check it: 3! = 3 × 2! says 6 = 3 × 2, and 2! = 2 × 1! says 2 = 2 × 1. Now take one more step down without changing the rule:
1! = 1 × 0!
We already know 1! is 1, so 1 = 1 × 0!. Exactly one number makes that true.
Not guessed: solved for. If the recursion is to reach the bottom rung at all, the value is forced.
The same move also shows where the factorials must stop. One more step down gives 0! = 0 × (−1)!, which would make 1 equal to 0 times something. No finite number can do that, so the ordinary factorial cannot be extended to the negative integers with finite values. The rule that hands us 0! = 1 is the same rule that closes the door below it.
Arranging nothing
The algebra forces 1. But what is that 1 actually counting?
Ask the original question literally: how many ways can you line up zero objects? The tempting answer is zero: no objects, no arrangements. That confuses the objects with the arrangements. An arrangement is a list, and there is exactly one empty list. Zero arrangements would mean the task is impossible, that there is no valid result you could hand back. But put down no objects and you are finished, correctly. The empty list is your answer. Do it again and you get the same empty list, not a second arrangement. The video animates this moment, and it is worth watching once.
The same idea is built into the binomial coefficients. The binomial coefficient “n choose k” counts the subsets of size k in a set of n things, and it is n! / (k! × (n − k)!). How many subsets with zero elements does a four-element set have? One: the empty set. The formula says 4! / (0! × 4!). Cancel the 4!s and you are left with 1 / 0!, which must equal 1. Look down the left edge of Pascal’s triangle: every 1 there counts a choice of zero elements. Any other value for 0! would make the factorial formula disagree with that entire edge.
Mathematicians put this as a general principle. A product of no numbers at all, an empty product, equals 1, the number that leaves everything unchanged when you multiply by it, in the same way that an empty sum equals 0. Zero factorial is the empty product.
The smooth curve through the factorials
So far we have counted whole objects. A stranger question: can a smooth curve pass through every factorial, filling in the gaps between whole numbers?
In the autumn of 1729, two mathematicians writing to Christian Goldbach answered yes. Daniel Bernoulli gave a formula in a letter dated 6 October 1729, and Leonhard Euler gave an infinite product in a letter of 13 October 1729. On 8 January 1730 Euler wrote again with an integral. Around 1811 Adrien-Marie Legendre gave the function its symbol, Γ, and rewrote Euler’s integral in the form used today:
Γ(x) = ∫0∞ tx−1 e−t dt, for positive x.
Read it as an area. Set x = 3 and the curve under the integral rises, peaks and decays towards 0; the area beneath it is Γ(3). One integration by parts turns the integral into the identity
Γ(x + 1) = x × Γ(x)
which is the factorial recursion again, now working even when x is not a whole number. And Γ(1) can be computed directly: the power of t becomes 0, leaving the area under e−t from 0 to ∞, which is exactly 1. That starting value and the recursion give Γ(n + 1) = n! for every whole number n ≥ 0. At n = 0, Γ(1) = 1 = 0!. The integral agrees with the counting, and it was never told the answer.
There is a catch. Smoothness alone does not pick out one curve; many smooth curves pass through the same points. In 1922 the Danish mathematicians Harald Bohr and Johannes Mollerup proved that Γ is the only function on the positive numbers that is positive, obeys the recursion, equals 1 at 1, and is log-convex, meaning its logarithm is convex. They published it in a textbook, believing it had already been proved. Our factorial curve is Γ shifted by one, which puts its value at 0 exactly at 1.
What a convention really is
People sometimes file 0! next to 0/0, as if both were arbitrary. They are opposites.
Division asks for a unique quotient. But 0 times any number is 0, so 0/0 picks out nothing. Limits show the trouble: as x → 0, x/x → 1, 2x/x → 2, and x/x² grows without bound as x → 0⁺. The same 0/0 shape allows incompatible answers, so the division stays undefined.
Zero factorial is the other extreme. The recursion, the empty arrangement, the binomial coefficients and the gamma integral all agree on 1, and they are connected, not four independent miracles. That is why their agreement matters.
Between the two sits 0⁰. In combinatorics and power series, defining it as 1 is useful and standard. But when a positive base and its exponent both approach 0, different paths can give different limits, so in limit problems 0⁰ remains an indeterminate form. Its value depends on the job.
A convention, then, is not a whim. It is a definition chosen to preserve the structure we are studying. For 0!, every structure gives the same answer. It quietly does its job elsewhere too: the series for the number e starts with 1/0! = 1, as told in why e is 2.718, and factorials count the shuffles in four letters, four envelopes.
One empty list, one recursion, one integral: 0! = 1.
