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For a function holomorphic on and inside a positively oriented simple closed contour $C$, the Cauchy integral formula givesIf an entire function is bounded by $M$, apply the derivative formula on a radius-$R$ circle about any $z$ to obtain $|f^{\prime }(z)|\le M/R$. Letting $R\to \infty$ makes every derivative $f^{\prime }(z)$ 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:The remainder is entire. The rational sum is $O(1/|z|)$ at infinity. If $n\ge 0$, this makes $g(z)=O(|z{|}^n)$; the Cauchy estimates at zero give $|g^{(m)}(0)|\le m!O(R^{n-m})$ for $m>n$, hence these derivatives vanish. Its Taylor series is a polynomial of degree at most $n$. If $n<0$, then $g\to 0$ at infinity, so it is bounded and equals zero by Liouville. In both cases, adding back the principal parts provesThis proof of polynomial growth forces a meromorphic function to be rational includes negative integers $n$, not only the polynomial-growth case $n\ge 0$.