Proof of $\frac{n^n}{3^n}\le n!$ for $\forall \ge6$ How to prove that $\frac{n^n}{3^n}\le n!$ for $\forall \ge6$ without using any calculus tools (without involving lim or Euler number).
I was told that the proof is similar to proving the definition of e, but I still cannot figure out how to do it.
Thank you for your help!
 A: Use induction. Just check it is true for $n=6$. Now suppose it is true for $n$, i.e.
$$\frac{n^n}{n!}\leq 3^n.$$ Consider
$$\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n}{n!}\leq\frac{3n^n}{n!}=3\cdot\frac{n^n}{n!}\leq 3\cdot 3^n=3^{n+1}$$
where the first inequality follows from 
$$\frac{(n+1)^n}{n^n}=\left(1+\frac{1}{n}\right)^n\leq e\leq 3.$$
A: In this answer, it is shown, using Bernoulli's Inequality that is proven there using induction, that
$$
\left(1+\frac1n\right)^{n+1}
$$
is a decreasing sequence. Therefore, for $n\ge5$,
$$
\left(\frac{n+1}{n}\right)^{n+1}\le\left(\frac65\right)^6=2.985984
$$
Thus, for $n\ge5$,
$$
\begin{align}
\left.\frac{(n+1)^{n+1}}{(n+1)!}\middle/\frac{n^n}{n!}\right.
&=\left(\frac{n+1}{n}\right)^n\\
&\lt\left(\frac{n+1}{n}\right)^{n+1}\\[9pt]
&\lt3
\end{align}
$$
Since
$$
\frac{5^5}{5!}\lt3^5
$$
we have that for $n\ge5$,
$$
\frac{n^n}{n!}\lt3^n\iff\frac{n^n}{3^n}\lt n!
$$
A: I think the proof you've been hinted towards is this one: applying the GM/HM inequality to $n+1$ copies of $\frac{n+1}{n}$ and one $1$ yields
$$ \Bigl(\frac{n+1}{n}\Bigr)^{(n+1)/(n+2)} 1^{1/(n+2)}
\ge \frac1{\frac{n+1}{n+2}\cdot\frac{n}{n+1} + \frac1{n+2}\cdot\frac11}
$$
Simplify and you will find that you have shown $\bigl(1+\frac1n\bigr)^{n+1}$ to be decreasing; then proceed as in robjohn's answer.
(This proof is dual to a standard one that $\bigl(1+\frac1n\bigr)^n$ is increasing, which is often given to justify the definition of $e$ as the limit.)
