I was just wondering while in the shower if the following limit

$\displaystyle \lim_{x\to{\infty}}\frac{e^{x}}{x!}=0$

Can be done another way...Please...Please don't put down the ways that are commonly seen...I know that

$\displaystyle x!$ increases much faster than $\displaystyle a^{x}$ thus the limit is 0...but does this method work?

For aesthetics n will denote ∞

So rewriting

$\displaystyle \lim_{x\to{\infty}}\frac{e^{x}}{x!}$

as

$\displaystyle \lim_{x\to\infty}\frac{1+x+\frac{x^2}{2!}+\text{.. .}+\frac{x^n}{n!}}{x(x-1)(x-2)\text{...}(x-(n-1))(x-n)}$

Factoring out an $\displaystyle x^{n}$ from top and bottom we get

$\displaystyle \lim_{x\to\infty}\frac{x^{n}\bigg[\frac{1}{x^{n}}+\frac{1}{2!x^{n-1}}+\frac{1}{3!x^{n-3}}+...+\frac{1}{n!}\bigg]}{x^n(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Cancelling we would get

$\displaystyle \lim_{x\to\infty}\frac{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{x^{n-2}2!}+...+\frac{1}{n!}}{(1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{n}{x})}$

Which would be equilvalent to

$\displaystyle \frac{0+0+0...+0}{(1-0)(1-0)...(1-0)}=\frac{0}{1}=0$

Is that mathematically sound?