How to prove that exponential grows faster than polynomial?

We can confine attention to $b \ge 1$. This is because, if $0<b<1$, then $n^b \le n$. If we can prove that $n/a^n$ approaches $0$, it will follow that $n^b/a^n$ approaches $0$ for any positive $b\le 1$. So from now on we take $b \ge 1$.

Now look at $n^b/a^n$, and take the $b$-th root. We get $$\frac{n}{(a^{1/b})^n}$$ or equivalently $$\frac{n}{c^n}$$ where $c=a^{1/b}$.

Note that $c>1$. If we can prove that $n/c^n$ approaches $0$ as $n\to\infty$, we will be finished.

For our original sequence consists of the $b$-th powers of the new sequence $(n/c^n)$. If we can show that $n/c^n$ has limit $0$, then after a while, $n/c^n \le 1$, and so, after a while, the old sequence is, term by term, $\le$ the new sequence. (Recall that $b\ge 1$.)

Progress, we need only look at $n/c^n$.

How do we continue? Any of the ways suggested by the other posts. Or else, let $c=1+d$. Note that $d$ is positive. Note also that from the Binomial Theorem, if $n \ge 2$ we have $$c^n=(1+d)^n \ge 1+dn+d^2n(n-1)/2\gt d^2(n)(n-1)/2.$$

It follows that $$\frac{n}{c^n}< \frac{n}{d^2(n)(n-1)/2}=\frac{2}{d^2(n-1)}.$$ and it is clear that $\dfrac{2}{d^2(n-1)} \to 0$ as $n\to\infty$.

We could prove this by induction on integers $k$:

$$ \lim_{n \to \infty} \frac{n^k}{a^n} = 0. $$

The case $k = 0$ is straightforward. I will leave the induction step to you. To see how this implies the statement for all real $b$, just note that every real number is less than some integer. In particular, $b \leq \lceil b \rceil$. Thus,

$$ 0 \leq \lim_{n \to \infty} \frac{n^b}{a^n} \leq \lim_{n \to \infty} \frac{n^{\lceil b \rceil}}{a^n} = 0. $$

The first inequality follows since all the terms are positive. The last equality follows from the induction we established previously.

First, notice that $(n+1)^b/n^b=1$ plus terms in negative powers of $n$, so it goes to 1 as $n\to\infty$ (with $b$ fixed). Now going from $n^b/a^n$ to $(n+1)^b/a^{n+1}$, you multiply the numerator by something that's going to 1 but the denominator by something that isn't (namely, by $a\gt1$).

You can expand as$$a^n = e^{n\log a} = 1+n\log a +\cdots +\frac{(n\log a)^b}{b!}+\cdots$$

$$\frac{a^n}{n^b} =\frac{1}{n^b}+\frac{1}{n^{b-1}}\log a+\cdots+\frac{(\log a)^{b}}{b!}+n\frac{(\log a)^{b+1}}{(b+1)!}+\cdots $$ So, $$\frac{a^n}{n^b}\geq n\frac{(\log a)^{b+1}}{(b+1)!}$$. This becomes arbitrarily large with large $n$

Therefore, $$\lim_{n\rightarrow\infty}\frac{a^n}{n^b} = \infty$$ Or $$\lim_{n\rightarrow\infty}\frac{n^b}{a^n} = 0$$

Regard the exponential function as the unique solution to the initial value problem $f(0) = 1$ with $f' = f$. The equivalent integral formulation is

$f(t) = 1 + \int_0^t f(s) ds$

First prove that $f$ is an increasing function. For $t \geq 0$, you can use a method of continuity argument by considering the set $\{ s ~:~ f \mbox{ increases on } [0, s] \}$. The supremum of this set cannot be finite. For $s < 0$ you can also consider the ODE satisfied by $f^2$, but you're not really concerned with that. Now that we know $f$ increase, we also know $f$ grows at least linearly:

$f(t) \geq 1 + \int_0^t 1 ds = (1 + t)$

It also grows at least quadratically..

$f(t) \geq f(\frac{t}{2}) + \int_{t/2}^t f(s) ds \geq (1 + t/2) f(t/2)$

And repeating yields

$f(t) \geq (1 + t/2)^2$

In general, $f(t) \geq (1 + t/n)^n$ by the same method. It's even true for $n$ not an integer. In this case, the function $(1 + t/n)^n$ is defined in the first place to be $e^{n \log( 1 + t/n) }$. Then, since log x is defined by the integral $\int_1^x \frac{1}{t} dt = x \int_0^1 (1 + sx)^{-1} ds $, we can rewrite

$(1 + t/n)^n = e^{n \log (1 + t/n) } = \mbox{exp}(t \int_0^1 (1 + \frac{s t}{n})^{-1} ds ) \leq e^t$

EDIT: This answer may not be so helpful to you. I didn't read the end of your question, so you may not know how to work with most of the things I've been writing about. The main reason why the exponential grows faster than a polynomial is because if $f$ is exponential, then $f(n+1)$ is at least a constant times $f(n)$, whereas when $f$ is a polynomial, $f(n+1)$ is roughly the same size as $f(n)$ when $n$ is large. After all, you can hardly tell the difference between the volume of a huge cube, and a huge cube 1 unit longer in each direction.