Proving $\frac{n^n}{3^n} < n! < \frac{n^n}{2^n}$ holds by induction

As a matter of style, I would start with $P(n)$, and from there try to prove $P(n+1)$. This makes it slightly less tricky to correctly manipulate inequalities.

So if we have $$\frac{n^n}{3^n} < n! < \frac{n^n}{2^n},$$ how do we get $P(n+1)$? Multiplying through by $n+1$ was a good instinct: $$\frac{(n+1)n^n}{3^n} < (n+1)! < \frac{(n+1)n^n}{2^n}.$$ There are now two pieces we need to prove: $$\frac{(n+1)^{n+1}}{3^{n+1}} < \frac{(n+1)n^n}{3^n}$$ and $$\frac{(n+1)n^n}{2^{n}} < \frac{(n+1)^{n+1}}{2^{n+1}}.$$

Here's a hint for these two inequalities: $\left(1+\frac{1}{n}\right)^n$ is an increasing function (prove this) that's bounded below by $2$ (when $n=1$) and converges to $e$. Let me know if you want something more explicit.

You have two inequalities to prove. You should prove them separately. There may be some duplication in typing, but the gain in logical clarity will be worth it.

Let $P(n)$ be the assertion that $n!<\frac{n^n}{3^n}$. We want to show that $P(n)$ holds for $n \ge 6$.

A minor computation deals with the base case $n=6$. Now I would like to show that if $P(n)$ holds when $n=k$, where $k$ is an integer $\ge 6$, then $P(n)$ holds when $n=k+1$. Unfortunately, you have chosen an example where this "induction step" is less easy than in many other cases.

So we assume that we know that $k! <\dfrac{k^k}{2^k}$. We want to show that $(k+1)!< \dfrac{(k+1)^{k+1}}{2^{k+1}}.$

How shall we exploit our knowledge about $k!$ to gain information about $(k+1)!$? It seems clear that we should try to exploit the relationship between $k!$ and $(k+1)!$. Since $(k+1)!=k!(k+1)$, and we know that $k! < \frac{k^k}{2^k}$, we have $$(k+1)!<(k+1)\frac{k^k}{2^k}.$$ If we could show that $(k+1)\dfrac{k^k}{2^k} <\dfrac{(k+1)^{k+1}}{2^{k+1}}$, we would be finished. A little algebra shows that "all" we need to do is to show that $k^k<\dfrac{(k+1)^k}{2}$, or equivalently that $\frac{(k+1)^k}{k^k}>2$. Rewrite this as the more familiar $\left(1+\frac{1}{k}\right)^k>2$.

Calculator experimentation seems to show that the required inequality is indeed true for all $k>1$. But we need to prove it, at least for $k \ge 6$. Induction anyone? We will use another tool.

Recall that by the Binomial Theorem, $$(1+x)^k =1+\binom{k}{1}x+\binom{k}{2}x^2 +\binom{k}{3}x^3+ \cdots +\binom{k}{k}x^k.$$ Put $x=\frac{1}{k}$. The first two terms in the binomial expansion of $\left(1+\frac{1}{k}\right)^k$ are $1$ and $k(1/k)$. They already add up to $2$. If $k>1$, the remaining terms put us above $2$. This finishes the proof.

An argument along similar lines shows that the other inequality also holds. We end up needing to show that $\left(1+\frac{1}{k}\right)^k <3$, which is somewhat harder than showing that $\left(1+\frac{1}{k}\right)^k >2$. There are a number of ways to handle that inequality. One way is to prove (by induction) that $\left(1+\frac{1}{k}\right)^k <3-\frac{1}{k}$. Strangely enough, this turns out to be easier to prove than the weaker inequality $\left(1+\frac{1}{k}\right)^k <3$.

Comment: Please note that the fact that you could not push the argument through does not mean that you don't know what induction is about. In your problem, the proof of the required inequalities is not obvious. You might try your hand at easier standard examples, such as the fact that $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$.