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 have to show that $$(n+1)!>\frac{(n+1)^{n+1}}{3^{n+1}}$$ multiplying $$n!>\frac{n^n}{3^n}$$ by $$n+1$$ we get 
$$(n+1)!>\frac{n^n(n+1)}{3^n}$$ this must be greater as $$\frac{(n+1)^{n+1}}{3^{n+1}}$$ this is true since $$\left(1+\frac{1}{n}\right)^n<3$$
