Proving $\frac{n^n}{3^n} < n!$ for $n\ge6$ by induction How would I prove this using mathematical induction:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$.
Here is what I have tried:
$\dfrac{n^n}{3^n} < n!$ for all $n \geq 6$
Base case: $\dfrac{6^6}{3^6} < 6!$. Yes, this is true.
Assume for $k$ it's true so $\dfrac{k^k}{3^k} < k!$.
Now prove for $n=k+1$, 
$\dfrac{(k+1)^{k+1}}{3^{k+1}} < (k+1)!$.
By assumption, we have that 
$\dfrac{k^k}{3^k} < k!$.
Multiplying by $k+1$, we get 
$$
\frac{k^k(k+1)}{3^k} < (k+1) k!
$$
and then
$$
\frac{k^k(k+1)}{3^k} < (k+1) !.
$$
I tried lots of things after this, but I am unable to figure out the proof.

We will use induction to show that $\;\;\color{red}{3^n n!>n^n}$ for all $\color{blue}{n\ge1}$:
1) This is true for $n=1$, since $3>1$.
2) Let $n\in\mathbb{N}$ with $3^n n!>n^n$.  Then $3^{n+1}(n+1)!=3(n+1)\big(3^n n!\big)>3(n+1)\big(n^n\big)$,
$\;\;\;$and $3(n+1)(n^n)>(n+1)^{n+1}\iff3n^n>(n+1)^{n}\iff3>\big(1+\frac{1}{n}\big)^n,\;$ and
$\hspace{.25 in}(1+\frac{1}{n})^n=\sum_{k=0}^{n}\binom{n}{k}(\frac{1}{n})^k=\sum_{k=0}^{n}\frac{n(n-1)(n-2)\cdots(n-(k-1))}{k!n^k}=\sum_{k=0}^{n}\frac{(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})}{k!}$
$\hspace{.87 in}\le\sum_{k=0}^{n}\frac{1}{k!}=2+\sum_{k=2}^n\frac{1}{k!}\le2+\sum_{k=2}^n\frac{1}{k(k-1)}=2+\sum_{k=2}^n(\frac{1}{k-1}-\frac{1}{k})=3-\frac{1}{n}<3$
$\;\;\;$Therefore $3^{n+1}(n+1)!>(n+1)^{n+1}$.
Thus $\color{red}{3^n n!>n^n}$ for all $\color{blue}{n\ge1}$ by induction.
