A simple proof about $e^x$? Do you guys think this is correct? I am trying to prove that there is no single-term polynomial function (oxymoron, I know) $f(x)$ which is always (or at least as x approaches infinity) greater than $e^x$ (I will try to expand this to any polynomial function later). Let $f(x)$ be such a function with the lowest possible degree. Then that means that the slope of $f$ will have to be greater than the slope of $e^x$ as x approaches infinity. This means that $f'(x)>e^x$ as x approaches infinity. However, we already assumed that $f(x)$ is the lowest degree which is greater than $e^x$, but $f'(x)$ is one degree lower than $f$. Therefore, this is a contradiction. What do you guys think? Is this valid? Thanks!
 A: For any polynomial $p(x)$, we have $\displaystyle\lim_{x\to\infty}\dfrac{p(x)}{e^x}=0$. This can be shown, for example, by using L'Hospital's Rule, and in other ways.
Your argument uses the derivative in a different way. The argument is somewhat informal, but it can be made formal. One could use induction. Let us prove by induction on degree that $\lim_{x\to\infty }\frac{p(x)}{e^x}=0$. 
We need to prove the degree $1$ base case. For the induction step, suppose the result holds for polynomials of degree $k$. We show it holds for polynomials of degree $k+1$. After a certain point $B$, by the induction hypothesis, we have $|p'(x)|\lt 1$. So $|p(B)+x)|\lt |p(B)|+x$. Thus the ratio $\frac{p(B+x)}{e^{B+x}}$ approaches $0$ as $x\to\infty$.
Remark: If we are allowed to use the power series expansion of $e^x$, there is an easy argument. Let $p(x)$ have degree $n$. For positive $x$, the power series expansion of $e^x$ shows us that $e^x\gt \frac{x^{n+1}}{(n+1)!}$. Now we can show easily that 
$$\lim_{x\to\infty}\frac{p(x)}{\frac{x^{n+1}}{(n+1)!}}=0.$$
A: I like this proof a lot. I think we can try to tighten it up a bit, though.
The main thing I would like you to justify is that "the slope of $f$ will have to be greater than the slope of $e^x$ as $x$ approaches infinity." In other words, there exists some $N\in \mathbb{R}$ such that $f'(x) \geq e^x$ for all $x\geq N$. Can you prove this?
Hint: Try supposing $f(x)\geq e^x$ for $x\geq M$, but $e^x \geq f'(x)$ for all $x\geq N$ and get a contradiction through integration.
