Proving $3^n < n!$ for some $n\in \mathbb{N}$ Prove:
There exists $N\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n\geq N$, we have $$3^n \leq n!$$  
This is what I did (I had to use a calculator).
Suppose $n\geq 7$. Then
$$n! = n(n-1)\ldots 7\cdot 6\cdot ... \cdot 1\\ \geq 7^{n-7}\cdot 7! \\ \geq 3^{n-7}\cdot 3^7 \\ = 3^n.$$
Hence, we have found $N$ such that whenever $n\geq N$, $3^n \leq n!$.  
I felt like I cheated a bit here... is there a better way to do this? I think I just have to show existence (non-constructive).
 A: 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: $n!\geq3^n$.
A: 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. 
A: We can write,
$$n!=n×(n-1)×(n-2)×(n-3)\cdots5×4×3×2×1$$

Now,
  $$(n-2)\ge 3$$
  $$(n-3)\ge 3$$
  $$(n-4)\ge 3$$
  $$\vdots$$
  $$5\ge 3$$
  $$4\ge 3$$

So,
$$n!\ge n(n-1)2×3^{n-4}$$
Now we want to find $n$ such that,
$$2(n-1)n\le 3^4$$
$$2(n-1)n\le 81$$
So that we can substitute $3^4$ in place of those terms,
This is simplified version of your problem

By hit and trial method, we get,$$n=7$$$$2(6)(7)\$$82\lt 81$$we can prove so by for next numbers by induction$$$$So for $n\ge 7$ ,$$n!\ge 3^4×3^{n-4}$$$$n!\ge 3^n$$

