I'm trying to prove that $\log(x!) = O(x\log x)$. I have got an idea but I'm not sure it's strong enough to be significant. I don't think I made any mistakes in reasoning, but I want to know if there is a path that leads to a more significant result.
First, with the exception of $0! = 1$, $x!$ can be rewritten as $x(x-1)(x-2)\cdot\cdot\cdot2\cdot1$.
Also, $x^x$ can be written (for nonnegative numbers and with the exception of $0^0$) as $x\cdot x\cdot \cdot \cdot x$.
We have that$,\text{ } \forall x$: $$x(x-1)(x-2)\cdot\cdot\cdot2\cdot1 < x\cdot x\cdot \cdot \cdot x$$ Therefore $x! = O(x^x)$. But, $x^x = e^{x\log x}$, therefore $x! = O(e^{x\log x})$. This leads to the conclusion that $\log(x!) = \log(O(e^{x\log x})) = O(x\log x).$
How does this look?
Just:
$\begin{align*} \log n! &= \sum_{1 \le k \le n} \log k \\ &= \sum_{1 \le k \le n} O(\log n) \\ &= n O(\log n) \\ &= O(n \log n) \end{align*}$
This is justified because it is a finite sum.