Prove that $\frac{x^n}{n!} < 1$ for $n$ sufficiently large

You are pretty spot on with your observation concerning the power series expansion of $e^x$; in fact, you are so nearly close to a proof that it's quite surprising you haven't finished it yourself! Because I believe that you'll learn more if you have to work out the appropriate gritty details, I'm only going to provide an outline, but I promise that the legwork on filling in the gaps will be worth it.

1) We know that $e^x$ has a power series expansion given by $\sum_{n=1}^\infty\frac{x^n}{n!}$.

2)By the ratio test, you can determine the radius of convergence of this series, finding that it converges on all of $\mathbb{R}$.

3) Since $\sum_{n=1}^\infty\frac{x^n}{n!}$ converges for all $x$, you know that the terms tend to $0$ in the limit as $n \rightarrow \infty$, and therefore are eventually smaller than any fixed positive constant you choose.

Hopefully that skeleton helps you flesh out the proof you are looking for!

Fix $x$ and let $n > 2\lceil x\rceil$, then $$\frac{x^n}{n!} \le \frac{\lceil x\rceil^n}{n!} = \color{blue}{\frac{\lceil x\rceil}{1}\frac{\lceil x\rceil}{2} \cdots \frac{\lceil x\rceil}{2 \lceil x\rceil}} \color{green}{\frac{\lceil x\rceil}{2\lceil x\rceil + 1} \cdots \frac{\lceil x\rceil}{n}}.$$ Notice that the product of the terms in $\color{blue}{\text{blue}}$ does not depend on $n$, while each of the terms in $\color{green}{\text{green}}$ is smaller than $\frac 12$. Then

$$\frac{x^n}{n!} \le \color{blue}{M(\lceil x\rceil)}\color{green}{\Big(\frac 12\Big)^{n - 2\lceil x\rceil}},$$ hence $$\lim_{n \to \infty}\frac{x^n}{n!} \le \lim_{n \to \infty}M(\lceil x\rceil)\Big(\frac 12\Big)^{n - 2\lceil x\rceil} = M(\lceil x\rceil)2^{2\lceil x\rceil} \lim_{n \to \infty}\Big(\frac 12\Big)^n = 0.$$

Note for $n$ even that $n! \geq (\frac{n}{2})^{\frac{n}{2}}$ since first $\frac{n}{2}$ terms of $n\cdot (n-1)...2\cdot 1$ are greater than or equal to $\frac{n}{2}$. Therefore since $n$ is eventually bigger than $2x^2$ it follows that $\frac{n}{2} > x^2$ and hence $(\frac{n}{2})^{\frac{1}{2}} > |x|$. Therefore $n! \geq ((\frac{n}{2})^{\frac{1}{2}})^{n} > |x|^n \geq x^{n}$.