# Why $0!$ is equal to $1$? [duplicate]

Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1$; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:

If $A$ is an associative and commutative ring, then we can define an unary operation on the set of all the finite subsets of our ring, denoted by $+ \left(A\right)$ and $\times \left(A\right)$. While it is intuitive to define $+ \left( \emptyset \right) =0$, why should the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?

• – wj32
Nov 1 '12 at 1:15
• In your question, where did this come from: "the product of zero number of elements be 1"? Nov 1 '12 at 11:02
• en.wikipedia.org/wiki/Empty_product Nov 1 '12 at 16:56

Here are two resons.

1. We can define $n!$ to be the number of rearrangements of $n$ distinct objects in a list. The empty list has one rearrangement: itself.

2. We can define $n!$ as the product of all positive integers $k$ with $1\le k \le n$. If $n$ is zero, we have an empty product. An empty product must be neutral for multiplication, so it must be 1.

• Nice attempt! But I think that I have an easier answer: "Because there is only one way to do nothing". Sep 6 '13 at 3:37

In general, we want the "associative" law to hold in the form: $$\times(A \cup B) = (\times A) (\times B)$$ whenever $A$ and $B$ are disjoint. What would this mean when $B = \emptyset$ ??

I think that the definition $0!=1$ is the one which makes most of the formulas work nicely.

For example $(n+1)!=n! (n+1)$ also works for $n=0$, as long as $0!=1$.

But most importantly, $\binom{n}{k}$ also works in the case $k=0$. Note that $\binom{n}{0}$ has to be $1$, if you want to have a nice formula for $(a+b)^n$. So for convenienceence, we define $\binom{n}{0}$ to be $1$, and if you want the binomial coefficient to be given by the standard formula, then $0!$ has to be 1.

Multiplying by $1$ is the same as not multiplying by anything at all.

As pointed out in one of the answers to this math.SX question, you can get the Gamma function as an extension of factorials, and then this falls out from it (though this isn't a very combinatorial answer).

The most elementary way to understand is just to work backwards. If you start with a sum of several elements and subtract them one at a time, you get $0$ after you finish. If you start with a product of several elements and divide them out one at a time, you get $1$ after you finish. Instead of thinking of $0!$ as an empty product and getting all metaphysical, just think of it as nonempty product with all the terms factored out.