Question

Why is `0! = 1` ?

Answer 1

We can do it by the definition of the factorial (assuming that `0! ne 0`, if that matters).

`N! = 1*2*3cdotsN`

Since `((N+1)!)/(N!) = (N!(N+1))/(N!) = N+1`, it follows that with `N = 0`,

`1 = (0!(0+1))/(0!)`,

and that

`0! = (0!(0+1))/1 = (1!)/1 = 1/1 = 1`.

Thus, `0! = 1`.

Answer 2

`0! = prod_(k=1)^0 k = 1`

Explanation:

The factorial of a non-negative integer is the product of all positive integers less than or equal to it.

We can write that as:

`n! = prod_(k=1)^n k`

If we apply this formula to `n=0` then we have:

`0! = prod_(k=1)^0 k = ?`

What we have here is an empty product - no terms multiplied together.

In the same way that an empty sum is `0` (the identity under addition), an empty product is `1` (the identity under multiplication).

So we can write:

`0! = prod_(k=1)^0 k = 1`