Why is $0! = 1$ $0! = 1$ ?

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Answer 1 · 2017-03-26T06:43:18

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

$N! = 1 \cdot 2 \cdot 3 \dots N$ $N! = 1*2*3cdotsN$

Since $\frac{(N + 1)!}{N!} = \frac{N! (N + 1)}{N!} = N + 1$ $((N+1)!)/(N!) = (N!(N+1))/(N!) = N+1$ , it follows that with $N = 0$ $N = 0$ ,

$1 = \frac{0! (0 + 1)}{0!}$ $1 = (0!(0+1))/(0!)$ ,

and that

$0! = \frac{0! (0 + 1)}{1} = \frac{1!}{1} = \frac{1}{1} = 1$ $0! = (0!(0+1))/1 = (1!)/1 = 1/1 = 1$ .

Thus, $0! = 1$ $0! = 1$ .

Answer 2 · 2017-03-26T07:59:09

$0! = 0 \prod k = 1 k = 1$ $0! = prod_(k=1)^0 k = 1$

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:

$n! = n \prod k = 1 k$ $n! = prod_(k=1)^n k$

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

$0! = 0 \prod k = 1 k = ?$ $0! = prod_(k=1)^0 k = ?$

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

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

So we can write:

$0! = 0 \prod k = 1 k = 1$ $0! = prod_(k=1)^0 k = 1$

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