prove that 2^n is O(n!)?

We say that f(x) is O(g(x)) iff there exist M and N such that

|f(x)| ≤ M|g(x)| for all x > N

So we want to show that there exist some M and N such that

2^n ≤ M n! for all n > N

I claim that M = 1, N = 3 works, and I will prove this by induction.

Note that 4! = 24 > 16 = 2^4, so it is true for n = 4.

Now suppose that for some n > 3, we have

n! > 2^n

Then (n + 1)! = (n + 1) n! > (n + 1) * 2^n (since n + 1 is positive)

Now n + 1 > 2 since n > 3, so (n + 1) * 2^n > 2 * 2^n = 2^[n + 1]

Hence if n! > 2^n for some n > 3, then (n + 1)! > 2^[n + 1] and hence, since 4! > 2^4, by the principle of mathematical induction, n! > 2^n for all n > 3.

So we have

2^n < n! for all n > 3, and hence 2^n is O(n!)

What does O(n!) stand for? I have not seen this symbolism before?