I'm having problem proving that $\log(n^n)=\Theta(\log(n!))$
I tried to use Stirling's formula but it seems it doesn't help me in this case.
This is what I wrote :
$$n \to \infty : \frac{\log(n!)}{\log(n^n)}=\frac{\log(\frac{\sqrt {2\pi n}}{e^n }.n^n)}{\log(n^n)}$$
Now what? nothing can be erased ... nothing can be made more simple ( Or maybe I don't know it)
Any idea?
I adapted my answer for you using an older Stack Overflow post:
$$\log(n!) = \log(1) + \log(2) + \dots + \log(n-1) + \log(n)$$
Upper bound can be calculated using Sterling's approximation:
$$\log(1) + \dots + \log(n) \leq \log(n) + \dots + \log(n) \text{ or } \boxed{n\log(n)}$$
Lower bound can be determined by reasoning: your answer will be the full expression after dropping the first half:
\begin{align*} {} \log(1) + \dots + \log(\frac{n}{2}) + \dots + \log(n) \\ {} \geq \log(\frac{n}{2}) + \dots + \log(n) \\ {} \geq \log(\frac{n}{2}) + ... + \log(\frac{n}{2}) \text{ or } \boxed{\frac{n\log(n/2)}{2}} \end{align*}