It's not really a proof, its more of a definition which turns out to be convenient in calculations, so we don't have to make exceptions in cases where "0!" might turn up.
If you let n! mean "the number of ways that n distinct objects can be ordered", you should ask: "in how many ways can 0 objects be ordered?". Reasonable answers would be 0 (if you have no objects, you cannot order them at all) or 1 (there is just one way, namely: don't
order them at all).
If we consider the the binomial coefficients (n choose k) we can look at (n choose 0), where n is a positive integer. This coefficient is defined as
n! / (n! 0!)
so if 0! were to be equal to zero, this would be undefined. If 0! = 1, then this would give n! / n! = 1.
Now the latter is more useful, for example, (n choose 0) means: "in how many ways can we choose 0 objects from a set of n?". The answer "1: namely, choose neither of them" is more sensible than "this question makes no sense, because (n choose 0) is undefined due to division by zero". Also, in expansions like
(a + b)^n = \binom{n}{0} a^n b^0 + \cdots = \frac{1}{0!} a^n + \cdots
it would make sense to have 0! = 1, because when you work out the brackets you will indeed get one and only one term with an. So we can write
(a + b)^n = \sum_{k = 0}^n \binom{n}{k} a^k b^{n - k}
where otherwise we would have to write
(a + b)^n = a^n + \left( \sum_{k = 1}^{n - 1} \binom{n}{k} a^k b^{n - k} \right) + b^n
(note that the final term, a0 bn, cannot be included in the sum either, because (n choose n) will contain 0! as well).
We can also use the recursive definition of the factorial to argue that 0! should be defined as being equal to 0. The function f(n) = n! can be recursively defined on the integers by
n! = n * (n - 1)!
where 1! = 1.
Although by definition the base case is 1! = 1, we could try and apply it to n = 1, and get
1! = 1 * 0!
If this is equal to 1, then 0! should be equal to 1.
[Note that going further would give 1 = 0! = 0 * (-1)!, making (-1)! undefined].
Again, this is not a rigorous argument, but it makes our life a little easier. There is no strong mathematical reason why we should be able to apply the recursion to n = 1 (after all, that's the definition), but it makes things a little more consistent if we do.
Finally there is the direct definition of n!, as the product of all numbers from 1 up to (and including) n. For example, 12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, 4! = 4 * 3 * 2 *1 and 1! = 1. If you know the following notation, it's simply
n! = \prod_{k = 0}^n k
Then 0! would be an "empty" product, which we usually define as having the value 1 (just like the empty sum has value 0).
So setting 0! = 1 is somewhat of an arbitrary choice (we could make n! be undefined for n = 0, or assign any other value), but it is a choice which allows us to use n! in great generality without having to make exceptions all the time, whenever n might possibly take on the value 0.
