Not an easy limit.

$\displaystyle \lim _{ n\rightarrow \infty }{ { \left[ \dfrac { n! }{ { n }^{ n } } \right] }^{ 1/n } } \\ \Rightarrow \displaystyle \lim _{ n\rightarrow \infty }{ { \left[ \dfrac { 1\times 2\times 3....n }{ n\times n\times n....n } \right] }^{ 1/n } } \\ \Rightarrow \displaystyle \lim _{ n\rightarrow \infty }{ { \left[ \dfrac { 1 }{ n } \times \dfrac { 2 }{ n } \times \dfrac { 3 }{ n } ....\dfrac { n }{ n } \right] }^{ 1/n } } \\ \therefore \displaystyle \log { A } = \displaystyle \lim _{ n\rightarrow \infty }{ \dfrac { 1 }{ n } { \left[ \log { \dfrac { 1 }{ n } } \times \log { \dfrac { 2 }{ n } } .....\log { \dfrac { n }{ n } } \right] } } \\ = \displaystyle \lim _{ n\rightarrow \infty }{ \displaystyle \sum _{ r=1 }^{ n }{ \dfrac { 1 }{ n } \log { \dfrac { r }{ n } } } } \\ = \displaystyle \int _{ 0 }^{ 1 }{ \log { x } } dx={ \left[ x\log { x } \right] }_{ 0 }^{ 1 }- \displaystyle \int _{ 0 }^{ 1 }{ x\times 1/x\quad dx } \\ =0- \displaystyle \int _{ 0 }^{ 1 }{ dx } =0-{ \left[ x \right] }_{ 0 }^{ 1 }\quad =\quad -1\\ \therefore \quad A={ e }^{ -1 }$

We can also use Cesaro's lemma with $u_n=\frac{n!}{n^n}$ then $\frac{u_{n+1}}{u_{n}}=(1-\frac{1}{n+1})^n$ which tends to $e^{-1}$ and we're done

The inequalities I often find useful are ( $\frac{n}{e}$ )^n <n!<( $\frac{n}{e}$ )^(n+1)
( $\frac{n}{e}$ )^n<n!<( $\frac{n}{e}$ )^(n+1)
These gives $\frac{n!^(1/n)}{n}$ →1/e
((1+ $\frac{1}{n}$ )^(1/n)<e<(1+ $\frac{1}{n}$ )^(n+1))