How to calculate $\lim\limits\_{x\to\infty} \frac{\log(x!)}{x\log(x)}$. Assume base $e$ (so $\ln)$. 

My attempt:

$$\lim\_{x\to\infty} \frac{\log(x!)}{x\log(x)}=\lim\_{x\to\infty}\frac{\log(1\cdot 2\cdot 3\cdots x)}{x\log x}=\lim\_{x\to\infty}\frac{\log(ax)}{x\log x}, a\gt 0$$

Applying LH rule:

$$\lim\_{x\to\infty} \frac{\frac{1}{x}}{\log(x)+1}=\lim\_{x\to\infty}\frac{1}{x(\log(x)+1)}=0$$

Wolfram tells me the answer is $1$. Where is my mistake?