Which has a higher order of growth, n! or n^n? 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?
 A: "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.
A: You are right that $n^n$ grows faster than $n!$. You can see this question with answer (thanks to @JohnHabert for his comment above) for a proof which says exactly what you are saying. This is illustrated by just considering the first couple of numbers
$$
\begin{align}
1^1 = 1 \quad &\quad\quad 1! = 1\\
2^2 = 4 \quad &\quad\quad 2! = 2\\
3^3 = 27 \quad &\quad\quad 3! = 6\\
4^4 = 256 \quad &\quad\quad 4! = 24 \\
5^5 = 3125 \quad &\quad\quad 5! = 120 \\
\end{align}
$$
(Your professor might just have misspoken. I would go ask him for clarification.)
A: If $n\in \mathbb{N}$, Then $n! = 1 \times 2 \times 3\times 4...............\times (n-2)\times (n-1)\times n$
Now $n\geq n$ and $n>(n-1)$ and $n>(n-2)$ and $n>(n-3)$......... $n>4\;\;,n>3\;\;,n>2\;\;,n>1$
So $n\cdot n\cdot n\cdot ............\cdot n(n-\bf{times})\geq n\cdot (n-1)\cdot (n-2)....3\cdot 2 \cdot 1$
So $n^n\geq n!$
A: Your professor is wrong, and the explanation can be found here on page 5.
