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.