Analytic function bounded by polynomial 
Prove that a function which is analytic in the whole plane and satisfies the inequality $|f(z)|<|z|^n$ for some $n$ and all sufficiently large $|z|$ reduces to a polynomial.

The function is analytic, so $f^{n}(z)$ exists for all $n$, all $z$. We have the Cauchy's integral formula for higher derivatives
$$f^{(n)}(a)=\frac{n!}{2\pi i}\int_C\frac{f(z)}{(z-a)^{n+1}}dz.$$
So $$\int_C\frac{f(z)}{(z-a)^{n+1}}dz<|z|^n$$ for all large $|z|$. How does that help?
 A: Use the integral formula for the $n+1^{\text{st}}$ derivative,
$$f^{(n+1)}(a) = \frac{(n+1)!}{2\pi i}\int_{\lvert z\rvert = R} \frac{f(z)}{(z-a)^{n+2}}\,dz.$$
Consider $\lvert a\rvert < 1$, and let $R \to \infty$. The standard estimate shows
$$\left\lvert f^{(n+1)}(a)\right\rvert \leqslant \frac{(n+1)!}{2\pi} \frac{R^n}{(R-1)^{n+2}}\cdot 2\pi R \leqslant \frac{C}{R},$$
so $f^{(n+1)} \equiv 0.$
A: Take $C$ to be a circle of radius $R$ around the origin, large enough to contain $a$. Parameterize via $z = Re^{it}$.  Then
\begin{align}
|f^{(n)}(a)| &= \frac{n!}{2\pi} \left|\int_C \frac{f(z)}{(z - a)^{n + 1}} dz \right| \\
&= \frac{n!}{2\pi} \left|\int_0^{2\pi} \frac{f(Re^{it})}{(Re^{it} - a)^{n + 1}} Rie^{it} dt\right| \\
&\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{|f(Re^{it})|}{|Re^{it} - a|^{n + 1}} R dt \\
&\le \frac{n!}{2\pi} \int_0^{2\pi} \frac{R^n}{|Re^{it} - a|^{n + 1}} R dt
\end{align}
The denominator can be bounded below by, say, $\frac{1}{2} R^{n + 1}$ by choosing $R$ sufficiently large, and so we see that
$$|f^{(n)}(a)| \le n! \frac{R^{n + 1}}{\frac{1}{2} R^{n + 1}} = 2n!$$
Hence, $f^{(n)}$ is a bounded entire function, and so is constant.
