Why does 0! is equal to 1 ?

When we have to calculate the factorial of a non-negative integer n, then it is denoted by n!, is the product of all positive integer less than or equal to n.

so,

𝑛!=𝑛∗(𝑛−1)∗(𝑛−2)∗(𝑛−3)∗...3∗2∗1$n!=n\ast (n-1)\ast (n-2)\ast (n-3)\ast ...3\ast 2\ast 1$

However, the recursive definition of factorial is of more use in this proof.

𝑛!={1𝑛∗(𝑛−1)!𝑛=0𝑛>0$n!=\{\begin{array}{ll}1& n=0\\ n\ast (n-1)!& n>0\end{array}$

The recursive definition of Factorial leads to one interesting way of expressing factorial numbers.

𝑛!=(𝑛+1)!(𝑛+1)$n!={\displaystyle \frac{(n+1)!}{(n+1)}}$

This is valid since, as we expand (𝑛+1)!$(n+1)!$ from the recursive definition, we can cancel (𝑛+1)$(n+1)$ term from both numerator and denominator to get 𝑛!$n!$. Or we can even calculate factorial in the numerator and then evaluate the division

For example,

5!=6!6=7206=

4!=5!5=1205

3!=4!4=244

2!=3!3=63

1!=2!2=22

In a similar way, if we try to express 0!$0!$ we get

0!=1!1=1

0!=1!1=1

And this ends our proof that  0!=1.

0!=1$1$.This proof is one of many ways, where. But this one is quite explanatory in itself.