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 ∗ ( n − 1 ) ∗ ( n − 2 ) ∗ ( n − 3 ) ∗ . . .3 ∗ 2 ∗ 1 However, the recursive definition of factorial is of more use in this proof. 𝑛 ! = { 1 𝑛 ∗ ( 𝑛 − 1 ) ! 𝑛 = 0 𝑛 > 0 n ! = { 1 n = 0 n ∗ ( n − 1 ) ! n > 0 The recursive definition of Factorial leads to one interesting way of expressing factorial numbers . 𝑛 ! = ( 𝑛 + 1 ) ! ( 𝑛 + 1 ) n ! = ( 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 = 720 6 = 4 ! = 5 ! 5 = 120 5 3 ! = 4 ! 4 = 24 4 2 ! = 3 ! 3