Horizontal asymptote is what you need.

$\begin{aligned} L&=\lim_{n\to\infty}\color{#3D99F6}\left(\frac{n!}{n^n}\right)^{\color{#D61F06}\frac{2n^4+1}{5n^5+1}}\\ \ln L &=\lim_{n\to\infty}\color{#D61F06}\frac{2n^4+1}{5n^5+1}\color{#3D99F6} \ln \left(\frac{n!}{n^n}\right)\\ &=\lim_{n\to\infty}\color{#D61F06}\frac{2n^4+1}{5n^5+1}\color{#3D99F6}\sum_{k=1}^{n}\ln \left(\frac{k}{n}\right)\\ &=\color{#D61F06}\lim_{n\to\infty}\frac{2n^5+n}{5n^5+1}\color{#3D99F6}\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln \left(\frac{k}{n}\right)\\ &=\color{#D61F06}\lim_{n\to\infty}\frac{2+\frac{1}{n^4}}{5+\frac{1}{n^5}}\color{#3D99F6} \int_{0}^{1}\ln x \,dx\\ &=\color{#D61F06}\left(\frac{2}{5}\right)\color{#3D99F6}\left(-1\right)\\ &=-\frac{2}{5}\\ \therefore \ln L &=-\frac{2}{5}\\ \implies L&=\boxed{\left(\frac{1}{e}\right)^{\frac{2}{5}}} \end{aligned}$