Proving $3^n < n!$ for some $n\in \mathbb{N}$

If $n!>3^n$ thus, $(n+1)!=(n+1)n!>(n+1)3^n>3\cdot3^n=3^{n+1}$ for all $n>2$.

Thus, it remains to make a base induction and $n=7$ is valid because $7!>3^7$.

Thus, for all $n\geq7$ we have: $n!\geq3^n$.

Generalize to $k$ for this result.

Note that if $b_n = k^n$ and $a_n = n!$, then $\frac{b_{n+1}}{a_{n+1}} = \frac{b_{n}}{a_{n}} \times \frac k{n+1}$. Since $\frac k{n+1} < 1$ for all $n > k$, it follows that $\frac {b_n}{a_n} \to 0$ (eventually monotonically) as $n\to \infty$, as $\frac{b_{n+l}}{a_{n+l}} \leq \frac{b_n}{a_n} \times \left(\frac{k}{n}\right)^l$ , hence is exponentially decreasing for large $n$. So, we have that for large enough $n$, $\frac{b_n}{a_n} < 1$, which is what you want for $k = 3$. This is a much stronger statement, though.