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Wouldn't this also mean that 1!=0!, why is this true?
Yes, precisely there is a unique function $\emptyset \to \emptyset$ (with empty graph), which happens to be a bijection ($\operatorname{id}_\emptyset$). Note, that $n!$ is the number of bijections $\{1,\dots, n\}\to \{1,\dots,n\}$.
For positive numbers the factorial function $n!$ is defined as the product of all positive integers less or equal to $n$. To define $0!$ we need to "extend" the definition. Another way to define it is to notice that:
$$(n-1)! = \frac{n!}{n}$$
Pluging $n=1$ we get: $0! = \frac{1!}{1} = 1$
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.