= Solution
For a function holomorphic on and inside a positively oriented simple closed contour $C$, the <Cauchy integral formula> gives
$$
f(z)=\frac1{2\pi i}\int_C\frac{f(\zeta)}{\zeta-z}\,d\zeta,\qquad f^{(m)}(z)=\frac{m!}{2\pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{m+1}}\,d\zeta.
$$
If an <entire function> is bounded by $M$, apply the derivative formula on a radius-$R$ circle about any $z$ to obtain $|f'(z)|\leq M/R$. Letting $R\to\infty$ makes every derivative $f'(z)$ zero, so $f$ is constant. This proves <Liouville theorem> directly.
For the given <meromorphic function>, the growth bound excludes poles outside the bounding disc. Its poles inside are finite in number, since poles cannot accumulate at a finite point of a meromorphic function. Subtract all their finite <principal parts>:
$$
g(z)=f(z)-\sum_p\sum_{j=1}^{m_p}\frac{a_{p,-j}}{(z-p)^j}.
$$
The remainder is entire. The rational sum is $O(1/|z|)$ at infinity. If $n\geq0$, this makes $g(z)=O(|z|^n)$; the <Cauchy estimates> at zero give $|g^{(m)}(0)|\leq m!\,O(R^{n-m})$ for $m>n$, hence these derivatives vanish. Its Taylor series is a polynomial of degree at most $n$. If $n<0$, then $g\to0$ at infinity, so it is bounded and equals zero by Liouville. In both cases, adding back the principal parts proves
$$
\boxed{f\text{ is a rational function}.}
$$
This proof of <polynomial growth forces a meromorphic function to be rational> includes negative integers $n$, not only the polynomial-growth case $n\geq0$.
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