Question: I'm trying to find an upper bound of upper bound $\frac{x^n}{n!}$ such that the $\lim\limits\_{n \rightarrow \infty}{upperBound} = 0$ for $n \in \mathbb N$ so I can use it in a squeeze theorem question. Now I was thinking I could use something like $\frac{1}{n!} <= \frac{1}{n}$, but I can't really come up with anything.

Answer: 
You could note that $e^x=\sum\_{n=0}^{\infty} \frac{x^n}{n!}$, that is, for every fixed $x$ it converges to $e^x$, thus as $n$ goes to infinity the $n$-th terms must go to zero, otherwise it would not be convergent.
Assume without restriction $x>0$, let $M \in N$ such that $M \geq x$. Let $k \in N$. We have
$$
\frac{x^{M+k}}{(M+k)!} = \frac{x^M}{M!}\cdot\frac{x}{M+1}\cdot\ldots\cdot\frac{x}{M+k} < \frac{x^M}{M!}\cdot(\frac{x}{M+1})^k \to 0
$$
Put differently, if $n\in N$, $n > M$, we have
$$
\frac{x^{n}}{n!} < \frac{x^M}{M!}\cdot(\frac{x}{M+1})^{n-M} \to 0
$$

Finally, if you want you can put $M=\text{ceil}(x)$.