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Prove $0! = 1$ from first principles

Why does $0! = 1$?

All I know of factorial is that $x!$ is equal to the product of all the numbers that come before it. The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0! = 0$. I'm perplexed as to why I have to account for this condition in my factorial function (Trying to learn Haskell). Thanks.

Ski Mask
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Brandon Frohbieter
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5 Answers5

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Mostly it is based on convention, when one wants to define the quantity $\binom{n}{0} = \frac{n!}{n! 0!}$ for example. An intuitive way to look at it is $n!$ counts the number of ways to arrange $n$ distinct objects in a line, and there is only one way to arrange nothing.

syxiao
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In a combinatorial sense, $n!$ refers to the number of ways of permuting $n$ objects. There is exactly one way to permute 0 objects, that is doing nothing, so $0!=1$.

There are plenty of resources that already answer this question. Also see:

Link

http://wiki.answers.com/Q/Why_is_zero_factorial_equal_to_one

http://en.wikipedia.org/wiki/Factorial#Definition

Glorfindel
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Eric Naslund
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  • Intending on marking as accepted, because I'm no mathematician and this response makes sense to a commoner. However, I'm still curious why there is 1 way to permute 0 things, instead of 0 ways. – Brandon Frohbieter Mar 06 '11 at 19:10
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    If you want to do nothing, there is a way to do it, you just don't do it. But if you say there are 0 ways to do nothing, then you are implying that it is impossible to do nothing, which is of course not the case. This is how I look at it. – Justin Mar 06 '11 at 19:37
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It's because $n! = \prod_{0<k\le n} k.$ ($n!$ is the product of all numbers $1, 2,\dots n$) For $n = 0$ there isn't any number greater then 0 and lesser or equal to $n$, so the product is empty; the empty product is defined by convention as 1.

FUZxxl
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It has many reasons.

For example, we can have a power series: $e^x = \sum_{n} x^n/n!$ and we would like the first term to be $1$.

Also, how many permutations are there of $0$ numbers? Well, one.

JT_NL
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We know that $\binom{n}{n}$ and $\binom{n}{0}= 1$. Thus $0! = 1$.

NebulousReveal
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    The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately. We treat binomial coefficients like $\binom{5}{6}$ separately already; the theorem assumes $0 \leq k \leq n$. – Carl Mummert Mar 06 '11 at 19:33