Find $\lim_{n \to \infty} \frac{n\ln(n)}{\ln(n!)}$ The answer is 
$$\lim_{n \to \infty} \frac{n\ln(n)}{\ln(n!)}=1$$
but I have no idea how to solve it. I tried to apply the L'hospitals rule since top and bottom tend to infinity, but $$\frac{d}{dn}\log(n!)=\frac{1}{n}+\frac{1}{n-1}...$$
and I don't know if that helps or not. Thanks for any hints/answers.
UPDATE: Thank you all for the answers, I am currently working through them to see if I understand it. I don't know why you all have been downvoted. It wasn't me!
 A: There is a standard (first year calculus) method for comparing the sum and integral of the same function, slightly different versions giving inequalites in both directions. I put a diagram for one of the directions below. The pair of inequalities found are a finite version of the "integral test" for infinite series.
$$ n (-1 + \log n ) = n \log n - n  = \int_1^n \; \; \log x \; \; dx < \sum_{j = 2}^n \log j \; = \log n!  $$
Shifting , we get
$$ \sum_{j = 2}^n \log j \; = \log n! < \int_1^{n+1} \; \; \log x \; \; dx  = (n +1) \log (n+1) - (n+1) = (n+1)(-1 + \log(n+1))$$
The reciprocal of the question,
$$  \frac{n (-1+\log n)}{n \log n}  < \frac{\log n!}{n \log n} < \frac{(n+1)(-1 + \log (n+1))}{n \log n}  $$
When $n > 1$ we get $n+1 < 2n,$ so $\log (n+1) < \log (2n) < \log 2 + \log n < 1 + \log n, $ so for $n > 1$ we have $-1 + \log(n+1) < \log n.$
$$ $$
$$  \frac{ (-1+\log n)}{ \log n}  < \frac{\log n!}{n \log n} < \frac{(n+1)\log n}{n \log n} = \frac{n+1}{n} $$
$$  $$
$$ 1 - \frac{ 1}{ \log n} \; \; < \; \; \frac{\log n!}{n \log n} \; \; < \; \; 1 + \frac{1}{n} $$
$$  $$


A: Note that by Stolz-Cesaro
$$\lim_{n \to \infty} \frac{n\ln(n)}{\ln(n!)}=\lim_{n \to \infty} \frac{(n+1)\ln(n+1)-n\ln n}{ln((n+1)!)-\ln n!}=\lim_{n \to \infty} \frac{\ln\left(1+\frac1n\right)^n+\ln (n+1)}{\ln(n+1)}$$
A: Use stolz cesaro. Let $a_n=\log (n^n)$ and $b_n=\log(n!)$. Then
$$
\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_n}=\lim_{n\to\infty}\frac{\log[(n+1)^{n+1}]-\log[n^{n}]}{\log(n+1)}=\lim_{n\to\infty}\frac{\log\left[(1+\frac{1}{n})^n\right]+\log(n+1)}{\log(n+1)}=1
$$
where we used the fact
$$
\left(1+\frac{1}{n}\right)^n\stackrel{n\to\infty}{\to} e.
$$
By Stolz Cesaro,
$$
\lim_{n\to\infty}\frac{a_n}{b_n}=1.
$$
A: For all $n>0$, it holds by elementary methods that (cf. here): $(n/3)^n\le n! \le n^n$. Hence
$$
1=\frac{n\log n}{\log n^n}\le \frac{n\log n}{\log n!} \le \frac{n\log n}{\log ((n/3)^n)}=\frac{n\log n}{n \log n-n\log 3}=1+o(1).
$$
A: Note that $\ln(n!) = \sum_{k=1}^n \ln(k)$. Now use that for any natural number $1 \leq j \leq n$
$$(n-n/j) \ln(n/j) \leq (n - \lceil n/j \rceil+1) \ln(\lceil n/j \rceil)  \leq \sum_{k=1}^n \ln(k) \leq n \ln(n).$$
Thus
$$1 \leq \liminf_{n \rightarrow \infty} \frac{n \ln(n)}{\ln(n!)} \leq \limsup_{n \rightarrow \infty} \frac{n \ln(n)}{\ln(n!)} \leq \frac{1}{1-j^{-1}}.$$
Letting $j \rightarrow \infty$ shows the claim.
