Question: Here is what I have so far:

Let $\varepsilon > 0$ be given
We want for all $n >N$, $|1/n! - 0|< \varepsilon$

We know $1/n <ε$ and $1/n! ≤ 1/n < ε$.
I don't know how to solve for $n$, given $1/n!$. And this is where I get stuck in my proof, I cannot solve for $n$, and therefore cannot pick $N > \cdots$.

Answer: 
Since $0 < \frac{1}{n!} < \frac{1}{n}$, by the squeeze theorem...

Without squeeze theorem: Let $\epsilon > 0$. Define $N = \epsilon^{-1}$. Then if $n > N$, $$\left|\frac{1}{n!}\right| = \frac{1}{n!} < \frac{1}{n} < \frac{1}{N} = \epsilon$$
How about this one: $$\sum\_{n=0}^{\infty}\frac{1}{n!}=e^1=e.$$ Thus the series converges and therefore $$\lim\_{n\to \infty}\frac{1}{n!}=0.$$