Q:

Show that $6^n/n! \le 6^5/5! \times 6/n$

I want to show that 
$$\frac{6^n}{n!} \le \frac{6^5}{5!} \cdot \frac 6n$$
without using induction, which I've done but is rather clunky. Is there a more straight forward way of doing this?

A:

For $n>5$
$$\frac{6^n}{n!} = \frac{6^6 6^{n-6}}{5!6\cdots n}=
\frac{6^6}{5!}\frac66\frac67\cdots\frac{6}{n-1}\frac1n
\le\frac{6^5}{5!}\cdot\frac 6n,$$
and you can check the cases $n=1,\dots,5$.

A:

$$\frac{6^n}{n!} = \prod_{i=1}^n \frac{6}{i}$$
Is increasing for $n<6$, constant for $n=6$ and decreasing $n>6$
Hence $$\frac{6^6}{6!} = \frac{6^5}{5!}$$ is a maximum.

