I need help to advance in the resolution of that limit: $ \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} $

$$\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{\frac{n+1}{n}}\right)^n = \frac{1}{(1+1/n)^n} \to1/e$$ as $n\to\infty$.

For reference, another approach starting from scratch:

Using the fact that $\ln n! = n\ln n - n +o(n)$ (see e.g. this proof of it), we can rewrite the quantities considered in the exponential form (almost always helpful) to obtain that $$ \sqrt[n]{\frac{n!}{n^n}} = e^{\frac{1}{n}\ln \frac{n!}{n^n}} = e^{\frac{1}{n}\left(\sum_{k=1}^n \ln k - n\ln n\right)} = e^{\frac{1}{n}\left(n\ln n - n +o(n) - n\ln n\right)} = e^{\frac{1}{n}\left(- n +o(n)\right)} = e^{-1 +o(1)} $$ so the limit is $e^{-1}$.

Let $A=\lim_{n\to\infty}\sqrt[n]{\dfrac{n!}{n^n}}$

$\ln A=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac rn$

Like The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$, this equals to $$\int_0^1\ln x\ dx$$