Here's an alternative proof. I'm certain that it's not the proof expected in an elementary class, as the tools it uses are a bit heavier. It has the advantage, though, of introducing the idea of conjugation and how it can be used to treat the point at $\infty$ - an important step towards extending polynomial dynamics to the dynamics of rational functions on the Riemann sphere.
The idea is to replace the polynomial $f(z)$ with the rational function $F(z)=1/f(1/z)$ and study the orbit of $F$. It turns out that $F$ has a super-attractive fixed point at zero and the statement about orbits of $f$ being attracted to $\infty$ can be recast in terms of orbits of $F$ being attracted to zero.
More precisely, let $f$ be a complex polynomial of degree at least two and define
$F(z) = \left\{\begin{array}{ll} 1/f(1/z) & z \neq 0 \\ 0 & z=0. \end{array}\right. $
Essentially, we've simply conjugated by the reciprocal function and filled the resulting removable discontinuity at the origin. $F$ is, in fact, differentiable at the origin with $F'(0)=0$, since
$F'(0)=\lim_{z\rightarrow 0} \frac{F(z)-F(0)}{z-0} = \lim_{z\rightarrow 0} \frac{1}{zf(1/z)}=0.$
That last equality is true because $f$ has degree at least two and, thus, $f(1/z)\rightarrow \infty$ faster than $z\rightarrow 0$.
Next, note that
$F(F(z)) = 1/f (1/(1/f(1/z))) = 1/f(f(1/z))$
and, more generally,
$F^n(z)=1/f^n(1/z).$
Substituting $z$ for $1/z$, we get
$F^n(1/z)=1/f^n(z).$
As a result, an orbit of $f$ generates an orbit of $F$ by taking the reciprocals of the terms in the orbit and vice-versa. Thus, an orbit in the neighborhood of $\infty$ for $f$ can be treated as an orbit in the neighborhood of $0$ for $F$.
We can now establish the result using the fact that zero is an attractive fixed point for $F$, for then there is an $r>0$ such that $|z| implies that $z$ converges to zero under iteration of $F$. Thus, if $|z|>1/r$, then $z$ converges to $\infty$ under iteration of $f$.
Note that the number $1/r$ becomes the crucial bound in the original question.