Help with finding Limit What is the limit  of  $$\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}}$$
Does any one can help, I am not sure how to solve this.

$$
\begin{aligned}
\lim _{n\to \infty }\left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}} &= \lim _{n\to \infty}\exp\left(\frac{3n^3+4}{4n^4-1}\ln\left(\frac{n!}{n^n}\right)\right)\\ &\approx
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}}{e^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\left(\frac{1}{2}\ln \left(2\pi n\right)-n\right)\right)\\ &=
\color{red}{e^{-\frac{3}{4}}}
\end{aligned}$$
Solved with Stirling approximation 
$$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x, \text{ for } x \to \infty$$
