Entire functions with $|f(z)/z^{n}|$ bounded at infinity are polynomials of degree at most $n$

Let \(f\) be entire with \(\abs{f(z)}\le M\abs z^n\) for \(\abs z\ge R\). Taylor expand at \(z=0\) to get \(f(z) = \sum_{j\geq 0}c_j z^j\) everywhere. Claim: \(c_{n+k} = 0\) for all \(k\geq 1\). By the formula for Taylor coefficients, it suffices to show \(f^{(n+k)}(0) = 0\) for all \(k\geq 1\). Apply the Cauchy estimate on the circle of radius \(\rho\geq R\): \[ \abs{ f^{(n+k)} (0)} &\leq {(n+k)! \over 2\pi} \int_{\abs{\xi} = \rho} \abs{f(\xi) \over \xi^{n+k+1}}\abs{\dxi}\\ &\leq {(n+k)! \over 2\pi} \int_{\abs{\xi} = \rho} {M\abs{\xi}^n \over \abs{\xi}^{n+k+1}}\abs{\dxi}\\ &= {(n+k)! \over 2\pi} {M\over \rho^{k+1}} \cdot 2\pi \rho \\ &= (n+k)!\,M\rho^{-k} \convergesto{\rho\to\infty} 0 .\] So \(f\) is a polynomial of degree at most \(n\).

If the bound holds on all of \(\CC\setminus\theset0\), then \(h(z) \da f(z)/z^n\) is bounded near \(0\), so its singularity at \(0\) is removable, and \(h\) is a bounded entire function. By Liouville \(h\) is constant, and \(f(z) = cz^n\).