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 $a\gt1$).