Since $ \begin{align*} & e^x = 1 + x + \ldots + \frac{x^n}{n!} + \frac{x^{n+1}}{(n+1)!} + \ldots , \\ & x^{-n}e^x \gt \frac{x}{(n+1)!} \rightarrow \infty \end{align*}$
when $ x \rightarrow \infty $. Hence $ e^x $ tends to infinity more rapidly than any power of $ x $.
"An Introduction to the Theory of Numbers" - G.H. Hardy and E. M. Wright
I understand the inequality but I do not see how this statement holds true or under what context it holds true. I may be misreading the statement and misinterpreting it. Obviously the inequality exists because the Power Series is a sum of terms of $ \frac{x^n}{n!} $ and so $ e^x $ will be greater than any single given term. How does this fact prove that $ e^x $ is tending more quickly towards infinity? And also faster than what exactly? If $ n $ is constant and $ x $ grows to infinity and $ \frac{e^x}{x^n} $ tends to infinity and not zero or one this tells me that $ e^x $ is growing larger than $ x^n $ and achieving a magnitude that is not of the same order . . . but what does the inequality have to do with anything? What is $ \frac{x}{(n+1)!} $ and how does it relate back to the issue at hand?