Prove that n! = O(n^n)

Question

How can I prove that n! = O(n^n)?

Answer 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 n! <= n^n .

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 k! < k^k , which proves that n! is indeed an element of O(n^n) .