how can we prove that zero factorial is equal to one?

Answer 1

Here's a proof: http://www.adonald.btinternet.co.uk/Factor/Zero.html

Answer 2

You can't prove it. That's just how it's defined.

However, there are many reasons it's defined that way. I will give two reasons that taking 0! = 1 is a good definition (assuming that you have already defined 1!, 2!, 3!, and so on). For example,

4! / 3! = 4; 3! / 2! = 3; 2! / 1! = 2. So we would expect 1! / 0! = 1. The only way to get this is to define 0! = 1.

For another example: n! measures the number of ways you can arrange n objects in a row. You can also think of this as the number of different orderings you can write a set with n elements.

There are 4! = 24 ways to write a set with 4 elements:
{1, 2, 3, 4}
{1, 2, 4, 3}
{1, 3, 2, 4}
{1, 3, 4, 2}
{1, 4, 2, 3}
...
{4, 3, 2, 1}

There are 3! = 6 ways to write a set with 3 elements:
{1, 2, 3}
{1, 3, 2}
{2, 1, 3}
{2, 3, 1}
{3, 1, 2}
{3, 2, 1}

There are 2! = 2 ways to write a set with 2 elements:
{1, 2}
{2, 1}

There are 1! = 1 ways to write a set with 1 element:
{1}

If a set has 0 elements, there is still 1 way to write it:
{}

So 0! should be 1.

Answer 3

The proof given above are about right. But it's mostly a matter of convenience.

However........ The Gamma Function (look it up in Wiki) is a generalization of the factorial function from the integers to the reals (actually, it's even defined for complex values, but you probably don't want to hear about that ☺) and the value of the Gamma Function at 0 --is-- 1.

Doug

Answer 4

0! = 1 is by definition of factorial, defined by induction,

n=0 : : 0! = 1

defined up to n!
then
(n+1)! = n! (n+1)

Thus 1! = 0! (1) = 1 AND SO ON ...

Answer 5

Binomial Theorem