问题
How can I prove that n! = O(n^n)?
回答1:
I assume that you want to prove that the function n! is an element of the set O(n^n). This can be proven quite easily:
Definition: A function f(n) is element of the set O(g(n)) if there exists a c>0 such that there exists a m such that for all k>m we have that f(k)<=c*g(k).
So, we have to compare n! against n^n. Let's write them one under another:
n! = n * (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1
n^n = n * n * n * n * ... * n * n * n
As you can see, the first line (n!) and the second line (n^n) have both exactly n items on the right side. If we compare these items, we see that every item is at most as large as it's corresponding item in the second line. Thus n! <= n^n (at least for n>5).
So, we can - look at the definition - say, that there exists c=1 such that there exists m=5 such that for all k>5 we have that k! < k^k, which proves that n! is indeed an element of O(n^n).
来源：https://stackoverflow.com/questions/4999448/prove-that-n-onn