What makes $0!$ equal to 1? 
Possible Duplicate:
Prove $0! = 1$ from first principles 

I don't understand how it's equal to 1. Also, I found that $(-x)!$ is equal to complex $\infty$. How is this so?
 A: Check out Factorial
One part of the answer given under 'Definition' is simply that it's convenient.  It allows things such as the Taylor series of $e^x$ to work.
Another reason is that it makes sense when viewing it from a permutation viewpoint, since $n!$ is the number of possible permutations of $n$ objects.  If we have 0 objects, there is only one possible permutation - that leaving all 0 objects where they are.
A: The equality $0!=1$ is by definition, for the convenience in various situations. Also it can be "proved" in some ways, for example we know that $\frac{n!}{n}=(n-1)!$, then let $n=1$ and $1=0!$, but still, I think that the right side must be defined before that. Then $-x!$ is not necessary equal to $\infty$, check http://en.wikipedia.org/wiki/Gamma_function. We have that $$\Gamma(n) = 1 \cdot 2 \cdot 3 \dots (n-1) = (n-1)!$$
and there you can find that it is undefined for negative integers, but defined for the rest of negative values http://upload.wikimedia.org/wikipedia/commons/b/b3/Gamma_function_2.png
A: The factorial $0! = 1$ by definition.
