It suffices to prove $\lim\limits_{n \to \infty}\dfrac{n\ln n}{\ln n!}=1$, which can be done as follows
\begin{align*} \lim_{n \to \infty}\frac{n\ln n}{\ln n!}&=\lim_{n \to \infty}\frac{n\ln n}{\sum\limits_{k=1}^n \ln k}\\ &=\lim_{n \to \infty}\frac{(n+1)\ln(n+1)-n\ln n}{\sum\limits_{k=1}^{n+1}\ln k-\sum\limits_{k=1}^{n}\ln k}\\ &=\lim_{n \to \infty}\frac{(n+1)\ln(n+1)-n\ln n}{\ln(n+1)}\\ &=\lim_{n \to \infty}\frac{\ln\left(1+\dfrac{1}{n}\right)^n+\ln(n+1)}{\ln(n+1)}\\ &=\lim_{n \to \infty}\left[\frac{1}{\ln(n+1)}\cdot \ln\left(1+\dfrac{1}{n}\right)^n+1\right]\\ &=0\cdot 1+1\\ &=1. \end{align*}
Please correct me if I'm wrong! THX!
Your key step is that $(n+1)\ln(n+1)-n\ln n =\ln\left(1+\dfrac{1}{n}\right)^n+\ln(n+1) $. If you know that $\left(1+\dfrac{1}{n}\right)^n $ is bounded, you are done.
This has been shown here (by me and others) a number of times, but it still has to be shown.
I assume that your steps in getting this were like this:
$\begin{array}\\ (n+1)\ln(n+1)-n\ln n &=n\ln(n+1)+\ln(n+1)-n\ln n\\ &=n(\ln(n)+\ln(1+1/n))+\ln(n+1)-n\ln n\\ &=n\ln(1+1/n)+\ln(n+1)\\ \end{array} $
If you know that $\ln(1+x) < x$ for $x > 0$, your result follows. The quickest proof of this is
$\ln(1+x) =\int_1^{1+x} \dfrac{dt}{t} =\int_0^{x} \dfrac{dt}{1+t} \lt\int_0^{x} dt =x$.