Prove that (5^n)/n! approaches 0 as n approaches infinity.
In this case it would be.
I think it is bleedingly obvious that
$\displaystyle \frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}$
must go to $\displaystyle 0$ since the denominator is so much bigger than the numerator for large values of $\displaystyle n$.
A simple application of the ratio test shows that $\displaystyle \displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0$
For reference, the ratio test says that $\displaystyle \displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}} {a_n}\bigg)$