Proof of $0! = 1$ I have been recently studying binomial theorem and in that we very frequently encounter factorials. But one queer thing which I found is $0!$. Even more queer is its value which is 

$0! = 1$.

I was told this fact and I accepted it(without any proof). But I am still confused about it.
How can we prove that $0!= 1 $?
 A: We can order $n$ elements in $n!$ ways. $0$ elements can be ordered in just one way.
One can choose $k$ of $n$ elements in ${n \choose k} = \frac{n!}{k!(n-k)!}$ ways. Obviously,
$$1 = {n\choose n} = \frac{n!}{n!\cdot 0!}\text{.}$$
Also, $0! = 1$ agrees with Euler's function $\Gamma$, for which $\Gamma(n) = (n-1)!$ holds.
These are not proves, but only good reasons why should $0!$ equal $1$.
A: As the other answers pointed out, $0! = 1$ is rather a definition than a thing we could prove, but there is some motivation behind it:


*

*First of all, it is a common convention, that the empty sum $\sum_\emptyset=0$ has value $0$. It seems to be a natural definition, because if we add something to the sum, we want to be able to sum by forming the disjoint union of the two sets: $$\sum_{x\in A} x + \sum_{x \in \emptyset} x = \sum_{x \in A \cup \emptyset} x = \sum_{x \in A}.$$
In the same way, we often define $\bigcup_\emptyset =\emptyset$ and $\prod_\emptyset = 1$. The latter gives us: $$n!:=\prod_{k=1}^nk \Rightarrow 0! = \prod_{k=1}^0 = \prod_\emptyset k = 1.$$ Thus, setting $0!$ is consistent to other conventions.

*Another answer comes from real analysis. By the Bohr-Mollerup Theorem, there is only one function $f:\Bbb R^+ \rightarrow \Bbb R$ that satisfies:


*

*$f(x+1)=xf(x) \forall x \in \Bbb R$

*$f$ is logarithmic convex

*$f$ interpolates the factorial in a sense that for $n \in \Bbb N$: $f(n+1) = n!$


This function is the Gamma-function and it has value $\Gamma(1) = 1$ corresponding to  $0! = 1$. Thus setting $0! = 1$ can also be considered to be a natural choice.
A: You can prove it from the definition of the factorial operation, which defines $0!=1$.  The question you should be asking is why we define it this way.
One argument in favor of this definition is that it preserves the important property that $$(n+1)! = (n+1)\cdot n!$$ even when $n=0$.  If we want to use this property to reason about factorials, we either need to accept that $0!=1$ or else be prepared to fill our arguments with special exceptions for $n=0$.
A different kind of argument proceeds from the objects that the factorial function is intended to model.  $n!$ is intended to be the number of ways of making an ordered list of $n$ distinct objects.  So for example $3!=6$ because there are six ways to order a list of three items:
$$\begin{array}{ccc}
(123) & (132) & (312) \\
(213) & (231) & (321) \end{array}$$
And $2! = 2$ because there are two ways to order a list of two items:
$$\begin{array}{cc}
(12) & (21)\end{array}$$
And $1!=1$ because there is one way to order a list of one item:
$$\begin{array}{c}
(1)\end{array}$$
And $0!=1$ because there is one way to order a list of zero items:
$$\begin{array}{c}
()\end{array}$$
$()$ is unquestionably a list of zero items, so why wouldn't you count it?
A: Note that:
$$n! = \frac{(n+1)!}{n+1}$$
So you can apply this for $0$. Actually it's quite nicely explained here
A: \begin{align}
4! & = 24. \quad \text{Divide by 4 and get:} \\
3! & = 6. \quad \text{Divide by 3 and get:} \\
2! & = 2. \quad \text{Divide by 2 and get:} \\
1! & = 1. \quad \text{Divide by 1 and get:} \\
0! & = 
\end{align}
Fill in the blank.
Not a proof, but this may shed some light.
Also, notice that when you multiply no numbers at all, you get $1$.  That's also been asked about here.  That's why $x^0=1$.
A: There is no proof for this statement, since $0! = 1$ is just a convenient and nice definition in many applications.
For instance, if you would like to think of $n!$ as the number of bijective maps from an $n$ element set to itself, then there is only one bijective map from an empty set to itself.
The other to interpret factorial is through the $\Gamma$ function defined as $\Gamma(z+1) = z \Gamma(z)$. If $z \in \mathbb{Z}^+ - \{1\}$, then $\Gamma(z) = (z-1)!$. Hence, if we would like to extend the notion of factorial beyond positive integers using the $\Gamma$ function, setting $z=1$, we get that $1! = 1 \cdot 0!$, which forces $0! = 1$.
