TITLE: solve $\ln(n!) = \Theta(n\ln(n))$ without stirling approximation
QUESTION [7 upvotes]: My homework was proving this equation which is simple using Stirling approximation. I was wondering if there is any other method to prove it - without Stirling -
I can prove $\ln(n!) = O(n\ln(n))$ like this:
$$\ln(n!)=\ln(n\cdot(n-1)\cdots2\cdot1)=\ln(n)+\ln(n-1)+\cdots+\ln(2)+\ln(1)≤n\ln(n)$$ which is obvious.
But I can't prove that $\ln(n!) = \Omega(n\ln(n))$.

REPLY [8 votes]: Use a multiplicative variant of Gauss's trick:
$$
(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots
                ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)
                \ge n^n
$$
This implies that $\ln(n!) \ge \dfrac12 n \ln n$.
The other direction is easy, as you mention, because $n! \le n^n$ and so $\ln(n!) \le n \ln n$.
So $\dfrac12 n \ln n \le \ln(n!) \le n \ln n$ and $\ln(n!) = \Theta(n\ln(n))$.