Is $\log(n!)=o(\log(n^n))$? This is a question from CLRS (3rd edition, Pg 61) :
I have to find out whether

$\log(n!) = o(\log(n^n))$

( Note it is Little-oh)
From this & this question, I can see why $\log(n!) = O(\log(n^n))$, however, I am having hard times proving the little oh.
The answer given here (CLRS Solutions - walkccc) is No meaning,

$\log(n!)$ is not $o(\log(n^n))$

Why is this true? My reasoning is that $\log(n^n)$ grows faster than $\log(n!)$, i.e. $\log(n!) < c \cdot \log(n^n)$ for all $c>0, n>n_0$, right?
 A: The answer is no. One way to see this is to notice that $n!>\left(\frac n2\right)^{n/2}$, because the $n/2$ largest factors are all at least $n/2$. (You have to be slightly more careful about what happens if $n$ is odd, but it's easy to check this works for $n>1$.)
Now this seems like a terrible bound as it is much lower than $n^n$, but when we take logs it turns out to be not that much lower. $\log(n^n)=n\log n$, and $$\log\left(\left(\frac n2\right)^{n/2}\right)=\frac{n\log n}{2}-\frac{n\log 2}{2}=n\log n\left(\frac12-\frac{\log 2}{2\log n}\right).$$
Since $\frac12-\frac{\log 2}{2\log n}\to \frac12$, we have $\log(n!)>\frac 13 \log(n^n)$ for $n$ sufficiently large.
(This bound is pretty weak, and you can get a limiting factor of $1$, instead of $1/2$, using Stirling's formula, but that isn't necessary to answer the question.)
A: $$\log(n!) - \log(n^n)=\left(\sum_{k=1}^n \log k\right)-n\log n = n\cdot \frac 1 n\sum_{k=1}^n\log\left(\frac k n\right) \simeq n \int_0^1 \log x dx$$
where we identify a Riemann sum for the last step.
So the answer is no.
A: $Log \, (n!)=\sum\limits_{k=1}^{n} Log \, (k) \geq \sum\limits_{k=[\sqrt n]}^{n} Log \, (k) \geq Log [\sqrt n] (n-[\sqrt n])$. So $\lim \inf \frac  1 {n log \, n}  Log \, (n!) \geq \frac 1  2$ proving that the ratio of the left side to the right side does noe tend to $0$.
