Why 0! = 1 and 1! = 0!

First, let's define factorial (!) or factorial notation (!). Factorial notation (!) is shorthand for multiplying consecutive descending natural numbers.

 n! = n(n-1)(n-2) ... 3•2•1

Let's solve,

 6! = 6*5*4*3*2*1 = 720 /6

 5! = 5*4*3*2"1 = 120 /5

 4! = 4*3*2*1 = 24 /4

 3! = 3*2*1 = 12 /3

 2! = 2*1 = 2 /2

 1! = 1 = 1 /1

 0! = ? = 1

It follows pattern wherein, from larger number factorial divided by the number, 6!/6, which descends to 0!.

From the pattern, 0! is obviously 1.

Therefore, 0! = 1.

Then, from the formula, n! = n(n-1)

With n = 1,

 1! = 1(1-1)!

 1! = 1(0)!

 1! = 0!

Therefore, 1! = 0!