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).

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: $n!\geq3^n$.
