does the series $\frac{\ln(n^{n})}{n!}$ converge?

Question

Does the series converge ?

My attempt:
$$\frac{\ln(n^{n})}{n!} = \frac{n \, \ln(n)}{(n-1)!} $$ and we known $\forall x \geqslant 0$, $$\ln(1+x)\leqslant x$$ and $$ \ln(n) = \ln(1+(n-1)) \leqslant n -1.$$

Answer 1

$$\ln\dfrac{n^n}{n!}=-n\cdot\dfrac1n\sum_{r=1}^n\ln\dfrac rn$$

Now $$\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac rn=\int_0^1\ln x\ dx$$

Answer 2

Without words:

$$n=8\to\frac{n^n}{n!}=\frac{8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}>\frac{8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8}{4\cdot4\cdot4\cdot4\cdot8\cdot8\cdot8\cdot8}=2\cdot2\cdot2\cdot2=2^{n/2}.$$

Answer 3

By ratio test

$$\frac{(n+1)\ln(n+1)}{(n+1)!}\frac{n!}{n\ln n}=\frac1n \frac{\ln(n+1)}{\ln n}\to 0$$