limits problem

Relevant wiki: Stirling's Formula

By Stirling's approximation for the factorial, we have $\frac{n!}{n^n} \sim \frac{\sqrt{2\pi n}}{e^n}$ and by an elementary argument $\frac{2n^4+1}{5n^5+1} \sim \frac{2}{5n}$

Since also the two-variable function $f(x,y) = x^y$ is continuous on $x>0$ , we have $\lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{\frac{2n^4+1}{5n^5+1}} = \lim_{n\to\infty} \left(\frac{\sqrt{2\pi n}}{e^n}\right)^{\frac{2}{5n}} = \left(\frac{1}{e}\right)^{\frac{2}{5}} \cdot \lim_{n\to\infty} (2\pi n)^{\frac{1}{5n}}$

This last limit can be evaluated using L'Hopital's rule as (writing $\exp(x) = e^{x}$ ) $\lim_{n\to\infty} (2\pi n)^{\frac{1}{5n}} = \lim_{n\to\infty} \exp\left(\frac{\ln(2\pi n)}{5n}\right) = \exp\left(\lim_{n\to\infty} \frac{\ln(2\pi n)}{5n}\right) = \exp\left(\lim_{n\to\infty} \frac{1/n}{5}\right) = \exp(0) = 1$

Therefore, the original limit is $\left(\frac{1}{e}\right)^{\frac{2}{5}} \cdot 1 = \boxed{\left(\frac{1}{e}\right)^{\frac{2}{5}}}$