Proving  $\frac{n^n}{3^n} < n! < \frac{n^n}{2^n}$ holds by induction I am Computer Science person, and I'm trying to beef up my mathematical skills a little bit.  This isn't homework and I don't have a professor I can go to to ask for help. I hope this is an appropriate forum for this type of question.
To learn more about inductive proofs, I am reading the Wikipedia article on them. However, I am having trouble solving one of the examples on the page:

$$P(n) :\quad \frac{n^n}{3^n} < n! < \frac{n^n}{2^n} .$$
  Prove $P(n) \; \forall n \in \mathbb{N}, n \ge 6$

Here is what I've done so far, with the "???" indicating where I don't know the next step.
Base case:
$$\begin{array}{rrcccl}
P(6)\colon& \frac{6^6}{3^6} &<& 6! &<& \frac{6^6}{2^6}\\
 \iff& 2^6 &<& 6! &<& 3^6\\
\iff& 64 &<& 720 &<& 729
\end{array}$$
Inductive step:
$$\begin{array}{rrccclc}
P(n+1) \colon&\frac{(n+1)^{n+1}}{3^{n+1}} &<& (n+1)! &<& \frac{(n+1)^{n+1}}{2^{n+1}}\\
\iff& \frac{(n+1)^{n+1}}{3^{n+1}} &<& (n+1)n! &<& \frac{(n+1)^{n+1}}{2^{n+1}}
&\text{Defn factorial}\\
 \iff& \frac{(n+1)(n+1)^n}{3 \cdot 3^n} &<& (n+1)n! &<& \frac{(n+1)(n+1)^n}{2 \cdot 2^n}
&\{a^{m+n} = a^m a^n\}\\
 \iff& \frac{(n+1)^n}{3 \cdot 3^n} &<& n! &<& \frac{(n+1)^n}{2 \cdot 2^n}
&\text{Divide by n+1}\\
 ???
\end{array}$$
Perhaps this just looks hopelessly foolish from the point of view of mathematicians, but I'm trying to learn.
Specifically what I am looking for is either guidance to arrive at the answer on my own, or pointers to documentation I should read to further my own understanding.

For the Inductive Step:
Suppose that we have
$$
\frac{n^n}{3^n} < n!
$$
and
$$
n! < \frac{n^n}{2^n}.
$$
We want to show that this implies that
$$
\frac{(n+1)^{n+1}}{3^{n+1}} < (n+1)!
$$
and
$$
(n+1)!  < \frac{(n+1)^{n+1}}{2^{n+1}}.
$$
Under these assumptions, to see the first inequality holds, write:
$$\begin{align}
\frac{(n+1)^{n+1}}{3^{n+1}} &= \frac{(n+1)}{3} \frac{(n+1)^n}{3^n} 
\\
&= \frac{n+1}{3} \left(\frac{n+1}{n}\right)^n \frac{n^n}{3^n}
\\
&< \frac{n+1}{3} \left(\frac{n+1}{n}\right)^n n! \qquad (\text{using inductive hypothesis})
\\
&= \left( 1 + \frac{1}{n} \right)^n \frac{1}{3} (n+1)!
\end{align}$$
and this reduces to showing that 
$$
\left( 1 + \frac{1}{n} \right)^n \frac{1}{3} < 1.
$$
The other inequality can be handled similarly.  You can probably save some paper by doing both inequalities at the same time, but if you are just starting out, I would suggest working with one inequality at a time.
