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Entire function bounded by a polynomial is a polynomial
Let $f \colon \mathbb{C}\to\mathbb{C}$ be an entire function (i.e. $f$ is analytic on $\mathbb{C}$) that satisfies $|f(z)|\leq M|z|^n$ where $n$ is a positive integer and $M \geq 0$ is a constant. By considering the estimates in the proof of Liouville's Theorem, deduce that $f$ has to be a polynomial of degree at most $n$.
$f$ is entire so it has an everywhere convergent taylor expansion $f(z) = \displaystyle\sum_{n=1}^{\infty }.$ For $k>n,$ $|a_k | = \frac{ f^{(k)} (0) }{k!} | = |\frac{1}{2\pi i} \displaystyle\int_{C_r} \frac{ f(\zeta )}{\zeta^{k+1} } | \le \frac{1}{2\pi } \frac{Mr^n}{2\pi r^{k+1} } 2\pi r = \frac{M}{r^{n-k} } $ - this follows from some basic integral inequalities and the Cauchy Integral Formula. Letting $r$ tend toward $\infty ,$ we have that $a_k=0.$
