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).

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. 
