TITLE: Which has a higher order of growth, n! or n^n?
QUESTION [8 upvotes]: In our algorithms class, my professor insists that n! has a higher order of growth than n^n. This doesn't make sense to me, when I work through what each expression means. 
n! = n * (n-1) * (n-2) * ... * 2 * 1
n^n = n * n * n * n * ... * n * n

Since n is, by definition, greater than n -1 or n-2, shouldn't any n^n, which is the product of n number of n's, be greater than n!, which is the product of an n number of integers less than n?

REPLY [6 votes]: "Higher order of growth" does not mean that $n!\lt n^n$ but the stronger property that 
$$n!/n^n\to0.$$
To prove that this property holds, note that $n\geqslant2k$ for every $1\leqslant k\leqslant n/2$ and $n\geqslant k$ for every $n/2\lt k\leqslant n$, hence
$$
n^n=\prod_{k=1}^{n}n\geqslant\prod_{1\leqslant k\leqslant n/2}(2k)\cdot\prod_{n/2\lt k\leqslant n}k=2^{n/2}\cdot n!,
$$
which proves that 
$$
\frac{n!}{n^n}\leqslant\frac1{2^{n/2}}\to0.
$$
Thus, indeed $n^n$ has a higher order of growth than $n!$, not the opposite.