# What makes $0!$ equal to 1? [duplicate]

Possible Duplicate:
Prove $0! = 1$ from first principles

I don't understand how it's equal to 1. Also, I found that $(-x)!$ is equal to complex $\infty$. How is this so?

-

## marked as duplicate by Alex Becker, Rahul, Kannappan Sampath, t.b., IlyaFeb 21 '12 at 15:07

See the answers here ... but did you really not find that question while researching yours? –  The Chaz 2.0 Feb 21 '12 at 14:12
We define empty products to be $1$ for the same reason that we define all empty sums to be zero. –  Thomas Andrews Feb 21 '12 at 14:23

Check out Factorial

One part of the answer given under 'Definition' is simply that it's convenient. It allows things such as the Taylor series of $e^x$ to work.

Another reason is that it makes sense when viewing it from a permutation viewpoint, since $n!$ is the number of possible permutations of $n$ objects. If we have 0 objects, there is only one possible permutation - that leaving all 0 objects where they are.

-
surely there are no permutations of arranging 0 objects? Eg if you have a table with no objects on, you can't arrange the no objects. Also you can't leave 0 objects where they are because, because the object(s) in 0 objects have no position? –  Jonathan. Feb 21 '12 at 15:14
If you don't like that, then maybe this one (also in the Wikipedia article) is better: Defining $0!=1$ allows a sensible formula for finding the number of ways to choose all the elements from a set of size $n$: $\binom{n}{n} = \frac{n!}{n!(n-n)!}$. $0!=0$ would not be useful here, but clearly the number of ways to choose all elements is $1$, so it's not a stretch to say $0!=1$ since then $\frac{n!}{n!0!} = \frac{n!}{n!} = 1$. The definition makes sense in a natural way. –  Alex Petzke Feb 21 '12 at 18:08
I don't see how "but clearly the number of ways to choose all elements is 1" is clear? –  Jonathan. Feb 21 '12 at 18:12
We're not choosing one element at a time. In general $\binom{n}{k}$ is the number of distinct subsets with $k$ elements out of a set with a total of $n$ elements. So when $k=n$, we find all subsets with $n$ elements, but the only subset with $n$ elements is the set itself. Hence $\binom{n}{k} = 1$. –  Alex Petzke Feb 21 '12 at 19:25
Just noticed... That last $\binom{n}{k}$ should be $\binom{n}{n}$. –  Alex Petzke Feb 21 '12 at 23:59

The equality $0!=1$ is by definition, for the convenience in various situations. Also it can be "proved" in some ways, for example we know that $\frac{n!}{n}=(n-1)!$, then let $n=1$ and $1=0!$, but still, I think that the right side must be defined before that. Then $-x!$ is not necessary equal to $\infty$, check http://en.wikipedia.org/wiki/Gamma_function. We have that $$\Gamma(n) = 1 \cdot 2 \cdot 3 \dots (n-1) = (n-1)!$$ and there you can find that it is undefined for negative integers, but defined for the rest of negative values http://upload.wikimedia.org/wikipedia/commons/b/b3/Gamma_function_2.png

-

The factorial $0! = 1$ by definition.

-