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 $n > N$, $$\left|\frac{1}{n!}\right| = \frac{1}{n!} < \frac{1}{n} < \frac{1}{N} = \epsilon$$
You don't need to *solve for* $n$, just to find *some* $n$ that is large enough that $1/n!$ is smaller than $\varepsilon$. It doesn't matter if you find an $N$ that is much larger than it *needs to be*, as long as it is finite.

By the hint that $1/n! \le 1/n$, therefore, it is enough to find an $N$ that is large enough that $1/n < \varepsilon$. And this is clearly the case if only $n > 1/\varepsilon$ ...