Why is 0! = 1 when 1! also equals 1

factorial definition:

$$\boxed{ n! = n \cdot (n-1)! }$$

Example:  

$$\\5! = 5 \cdot 4! \\ 4! = 4 \cdot 3! \\ 3! = 3 \cdot 2! \\ 2! = 2 \cdot 1! \\ 1! = 1 \cdot 0! \\ 0! = 1 ~ \text{(convention)}\\ \boxed{\,5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot \textcolor[rgb]{150,0,0}{ 1 } \,}$$

There are several answers to this.

I could say it is by convention.

or I could say because it works.

Think about this.    3! is the number of ways you can line 3 people up in a queue

2! is the number of ways that you can line up 2 people    A then b  or B then A

1! is the number of ways you can line up 1 person.

How many ways can you line up 0 people.  Well 1 way.  There is only one way that you can put nobody in a queque

0!=1

I am sure that there are many more explanations but I guess the most important one is that it makes probablility questions work.  

Lame perhaps but true :)