Suppose that $f:\mathbb{C}\to \mathbb{C}$ is an entire function, $n\ge 1$ is an integer, and that $|f(z)|\le (1+|z|)^n$ for all $z\in\mathbb{C}$. Prove that $f$ is a polynomial.
Proof:
Since $f$ is entire, we can write it as $$f(z) = \sum\limits_{m=0}^\infty c_m z^m$$
By Cauchy Estimates, for any closed disk $\mathbb{D}$ of radius $r>0$ centred about $z=0$, $$|f^{(m)}(0)|\le \frac{M_r m!}{r^m}$$ where $M_r=\sup\{|f(z)|:z\in\mathbb{D}\}\le (1+r)^n$. Let $n<m$, then
$$ |c_m|=\frac{|f^{(m)(0)}|}{m!}\le \frac{M_r m!}{m! r^m}=\frac{M_r}{r^m}\le \frac{(1+r)^n}{r^m} $$
Also, $|c_m|=\lim\limits_{r\to \infty}|c_m|\le 0$.
Hence, $f(z)=c_0+c_1z+\dots+c_nz^n$ is a polynomial of at most degree $m$.
My question is this: how does the above proof show that $f$ is a polynomial?
From Cauchy's Integral Formula, for an analytic function $f$ we have
$$f^{(m)}(0)=\frac{m!}{2\pi i}\oint_{|z|=r>0}\frac{f(z)}{z^{m+1}}\,dz$$
Given that $|f(z)|\le (1+|z|)^n$, we have the estimate for $f^{(m)}(z_0)$
$$|f^{(m)}(0)|\le \frac{m!}{2\pi }\frac{2\pi r (1+r^n)}{r^{m+1}}$$
For $m>n$, we find that
$$f^{(m)}(0)=0$$
Thus, we have
$$\begin{align} f(z)&=\sum_{m=0}^\infty\frac{f^{m}(0)}{m!}z^m\\\\ &=\sum_{m=0}^n\frac{f^{m}(0)}{m!}z^m+\underbrace{\sum_{m=n+1}^\infty\frac{f^{m}(0)}{m!}z^m}_{=0}\\\\ &=\underbrace{\sum_{m=0}^n\frac{f^{m}(0)}{m!}z^m}_{\text{A polynomial of order}\,n} \end{align}$$
Hence, $f$ is a polynomial of, at most, order $n$.