Facts-1

Suppose we take a to be some whole number.
$a!=\frac{(a+1)!}{a+1}$
So, $0!=\frac{(0+1)!}{0+1}=\boxed 1.$

$n!$ is the number of ways we can arrange $n$ objects. If we have $0$ objects we can only arrange it in one way that is to put $0$ object. This may seem a bit philosophical and eccentric :P. An example:- How can you arrange 3 pens A,B,C? The answer is 6 because we can arrange them as ABC, ACB, BAC, BCA, CAB, CBA. How many ways can we arrange 2 pens A and B? The answer is 4:- AB and BA. How can ways can we arange 1 pen A? Only in 1 way that is A. Now, how many ways can you arrange 0 pens. There is one way! Dont place anything at all. That is the way of arranging 0 pens. So, we get $\color{#3D99F6}{\boxed{0! = 1}}$ .

Another proof is using nC0 that is $n \choose 0$ . nC0=1 and it is equal to $\frac {n!}{n!0!}$ . From here we can conclude 0! Is 1.