Wouldn't this also mean that 1!=0!, why is this true?
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One definition of the factorial that is more general than the usual
$$ N! = N\cdot(N-1) \dots 1 $$
is via the gamma function, where
$$ \Gamma(N) = (N-1)! = \int_0^{\infty} x^{N-1}e^{-x} dx $$
This definition is not limited to positive integers, and in fact can be taken as the definition of the factorial for non-integers. With this definition, you can quite clearly see that
$$ 0! = \Gamma(1) = \int_0^{\infty} e^{-x} dx = 1 $$
If you are starting from the "usual" definition of the factorial, in my opinion it is best to take the statement $0! = 1$ as a part of the definition of the factorial function, as anything else would require proofs using the factorial to include special cases for $0!$ and $1!$. It's a definition that is consistent and makes our lives easier.
There is exactly one permutation of a set of zero elements.
Mostly it is based on convention, when one wants to define the quantity $\binom{n}{0} = \frac{n!}{n! 0!}$ for example. An intuitive way to look at it is $n!$ counts the number of ways to arrange $n$ distinct objects in a line, and there is only one way to arrange nothing.
In a combinatorial sense, $n!$ refers to the number of ways of permuting $n$ objects. There is exactly one way to permute 0 objects, that is doing nothing, so $0!=1$.
There are plenty of resources that already answer this question. Also see:
Link
http://wiki.answers.com/Q/Why_is_zero_factorial_equal_to_one
http://en.wikipedia.org/wiki/Factorial#Definition