First recall the famous limit formula for the constant $e$:
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e\approx 2.7182 \ldots<3.\tag{1}$$
Now recall the binomial theorem:
Binomial theorem: $$(x+y)^n=\sum_{i=0}^n\binom{n}{i}x^{i}y^{n-i}.\tag{2}$$
We may use $(2)$ (with $x=1/k$, $y=1$, and $n=k+1$) to rewrite $\sum_{i=0}^{k+1}\binom{k+1}{i}k^{1-i}$ in the following manner (rewriting this sum is necessary later on in the induction proof):
\begin{align}
\sum_{i=0}^{k+1}\binom{k+1}{i}k^{1-i}&= \sum_{i=0}^{k+1}\binom{k+1}{i}(1/k)^ik\tag{$k^{1-i}=(1/k)^ik$}\\[1em]
&= k\sum_{i=0}^{k+1}\binom{k+1}{i}(1/k)^i\tag{pull out constant $k$}\\[1em]
&= k\left(\frac{1}{k}+1\right)^{k+1}\tag{by $(2)$}\\[1em]
&= k\left(\frac{1}{k}+1\right)\left(\frac{1}{k}+1\right)^k\tag{manipulate}\\[1em]
&= \left(1+\frac{1}{k}\right)^k(k+1).\tag{simplify}
\end{align}
Hence, we have that
$$
\sum_{i=0}^{k+1}\binom{k+1}{i}k^{1-i}=\left(1+\frac{1}{k}\right)^k(k+1).\tag{3}
$$
With this in mind, see if you can follow the proof below by induction.
Claim: $\frac{n^n}{3^n}<n!$ for all $n\geq 6$.
Proof. For any integer $n$, let $S(n)$ denote the statement
$$
S(n) : n^n<3^nn!.
$$
Base case ($n=6$): $S(6)$ says that $46656=6^6<3^66!=524880$, and this is true.
Inductive step $S(k)\to S(k+1)$: Fix some $k\geq 6$. Assume that
$$
S(k) : \color{blue}{k^k<3^kk!}
$$
holds. To be shown is that
$$
S(k+1) : \color{green}{(k+1)^{k+1}<3^{k+1}(k+1)!}
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\color{green}{(k+1)^{k+1}}&= \sum_{i=0}^{k+1}\binom{k+1}{i}k^{k+1-i}\tag{by $(2)$}\\[1em]
&= \color{blue}{k^k}\left(\sum_{i=0}^{k+1}\binom{k+1}{i}k^{1-i}\right)\tag{factor out $k^k$}\\[1em]
&\color{blue}{<} \color{blue}{3^kk!}\left(\sum_{i=0}^{k+1}\binom{k+1}{i}k^{1-i}\right)\tag{by $S(k)$}\\[1em]
&= 3^kk!\left(1+\frac{1}{k}\right)^k(k+1)\tag{by $(3)$}\\[1em]
&< 3^kk![3(k+1)]\tag{by $(1)$}\\[1em]
&= \color{green}{3^{k+1}(k+1)!},\tag{by definition(s)}
\end{align}
one arrives at the right-hand side of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 6$. $\blacksquare$
