TITLE: How to prove that exponential grows faster than polynomial?
QUESTION [96 upvotes]: In other words, how to prove:

For all real constants $a$ and $b$ such that $a > 1$,
$$\lim_{n\to\infty}\frac{n^b}{a^n} = 0$$

I know the definition of limit but I feel that it's not enough to prove this theorem.

REPLY [2 votes]: Rudin states this as a theorem in his Principle of Mathematical Analysis. See Theorem 3.20 (d) on page 57 of the 3rd edition. 
This theorem is in Chapter 3 Numerical Sequences and Series of the book, way before the definition of derivatives and even before the discussion of continuous functions. Rudin's presentation is very succinct and reads like a poem:

Theorem 3.20 (d). If $p>0$ and $\alpha$ is real, then $\displaystyle\lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0$. 

Proof. Let $k$ be an integer such that $k>\alpha$, $k>0$. For $n>2k$,
$$
(1+p)^n>\binom{n}{k}p^k=\frac{n(n-1)\cdots(n-k+1)}{k!}p^k>\frac{n^kp^k}{2^kk!}.
$$
Hence
$$
0<\frac{n^\alpha}{(1+p)^n}<\frac{2^kk!}{p^k}n^{\alpha-k}\quad (n>2k).
$$
Since $\alpha-k<0$, $n^{\alpha-k}\to 0$, by (a).†

† Here (a) is a statement in Theorem 3.20, which says that 

if $p>0$, then $\displaystyle \lim_{n\to\infty}\frac{1}{n^p}=0$. 

The proof of this fact in Rudin is of one line: Take $n>(1/\varepsilon)^{1/p}$. The author also makes a comment that "Note that the archimedean property of the real number system is used here."