How to prove that $f(x)=e^{x}$ is not a polynomial. It is obvious that if we differentiate $f(x) = e^{x}$ with respect to x we will get again and again $e^{x}$. Can we conclude anything by considering the behavior at $\pm\alpha$
 A: If $f(x)$ is a polynomial then
$$f(x)=a_nx^n+...+a_0$$
and $f^{(n+1)}(x)=0$, which is a contradiction because
$$f^{(k)}(x)=e^x>0$$
for all $k$
A: Without using differentiation, but with using limits:

Suppose $f(x)$ is a polynomial such that for all $x \in \mathbb{R}$, $f(x)=e^x$. 
Let the degree of $f$ be $n \in \mathbb{N}$. Let the cofficient of $x^n$ be $a$. Since this is the leading coefficient, it must be nonzero. 
Since $n\in\mathbb{N}$, $n$ must be even or odd. 


*

*If $n$ is odd and $a$ is positive, then $$\lim_{x\to-\infty}  f(x) = \infty$$

*If $n$ is odd and $a$ is negative, then $$\lim_{x\to-\infty}  f(x) = -\infty$$

*If $n$ is even and $a$ is positive, then $$\lim_{x\to-\infty}  f(x) = \infty$$

*If $n$ is even and $a$ is negative, then $$\lim_{x\to-\infty}  f(x) = \infty$$
We know that
$$\lim_{x\to -\infty} e^x = 0$$
This forces us to conclude that $a$ is neither positive nor negative, which means $a$ must be zero. That's a contradiction. 
A: One can use the fact that $x^{n} /e^{x} \to 0$ as $x\to\infty$ for any $n$. If $p(x) $ is polynomial of degree $m$ choose $n=m+1$ and then $x^{n} /p(x) \to \pm\infty$ and hence $p(x) \neq e^{x} $. The same limit property of $e^{x} $ can be used to prove that $e^{x} $ is not an algebraic function. See this answer to a related question.
