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Prove $0! = 1$ from first principles
Why does 0! = 1?
If I'm right, factorial $!$ means:
$$n!=1 \cdot 2 \cdot 3 \cdot 4 \cdots n $$
so:
$$ \begin{align} 5!&=1\cdot2\cdot3\cdot4\cdot5=120\\ 4!&=1\cdot2\cdot3\cdot4=24\\ 3!&=1\cdot2\cdot3=6\\ 2!&=1\cdot2=2\\ 1!&=1 \end{align} $$
But what is $n!$ when $n=0$?
It can't be undefined and it can't be $n!=0$, since those are illegal in known equations like:
$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$
So what is it?
On
It's conventionally defined as $0! = 1$. This agrees with the gamma function $\Gamma(1) = (n-1)! = 1$.
On
The number of ways to permute a set of $n$ objects is $n!$ (including the identity permutation). Your question can be reinterpreted in the following way: How many ways can one permute the elements of the empty set? Since the question can be viewed as ill-formed, one answers by conventionally defining the number to be $1$. That is, $0! = 1$. Compare this to computing the number of maps from a set of $m$ elements to a set of $n$ elements, $n^{m}$, in the case both sets are empty. Again, $0^{0} = 1$.
On
We know $\binom n r=\frac{n\cdot(n-1)\cdots (n-r+1)}{r!}=\frac{n!}{r! (n-r)!}$ ---> the number of ways we can choose $r$ elements from $n$ elements.
If $r=n,$ we can take $n$ elements from $n$ elements in $\frac{n\cdot(n-1)\cdots (n-n+1)}{n!}=1$ way.
So, $1=\frac{n!}{n! (n-n)!}=\frac 1{0!}\implies 0!=1$
Similarly, if $r>n, \binom n r=\frac{n\cdot(n-1)\cdots (n-r+1)}{r!}=0\implies \frac1 {(n-r)!}=0 \implies \frac 1{s!}=0$ if $s=n-r<0$
On
$0!=1$.
Reason 1: $(n-1)!=\dfrac{n!}n$, so $0!= \frac{1!}1$.
Reason 2: $n!$ is the number of bijections of a set of cardinality $n$. The only set of cardinality $n$ is the empty set, the number of functions from the empty set to the empty set is 1.
Reason 3: $n! = \int_0^{\infty} x^ne^{-x}dx$. The value for $n=0$ is 1 (and this is actually used as base of the induction proof).
$0! = 1$ is consistent with, and for reasons related to, how we define the empty product. See this entry on empty product.
Some of the most common examples are the following:
Just as ${n^0 = 1}$ for any $n$, we define, as a convention, $0!$ to be $1$.
Added observation:
$$e^x = 1 + \frac {x} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... = 1 + \sum_{n=1}^\infty \frac{x^n}{n!} \tag{1}$$
But the following is a more concise definition: $$e^x = \frac {x^0} {0!} + \frac {x^1} {1!} + \frac {x^2} {2!} + \frac {x^3} {3!} + ... = \sum_{n=0}^\infty \frac{x^n}{n!}\tag{2}$$
$(1)$ and $(2)$ are equal if and only if $$\;\;\displaystyle e^0 = \frac{x^0}{0!} = \frac {1}{0!} = 1 \iff 0! = 1.$$