$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\frac{(n+1)!5^{n}}{n!5^{n+1}}=\lim_{n\to \infty}\frac{n+1}{5}=\infty,$$ thus $\lim_{n\to \infty}a_n=\infty$ so that we cannot possibly have $\frac{n!}{5^n}\leq C$ for a $C\in\mathbb{R}$ as $n\to\infty$.
Just observe that, using $a_n=\frac{n!}{5^n}$,
$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to \infty}\frac{(n+1)!5^{n}}{n!5^{n+1}}=\lim_{n\to \infty}\frac{n+1}{5}=\infty,$$ thus $\lim_{n\to \infty}a_n=\infty$ so that we cannot possibly have $\frac{n!}{5^n}\leq C$ for a $C\in\mathbb{R}$ as $n\to\infty$.