Proving an Entire Function is a Polynomial I had this question on last semesters qualifying exam in complex analysis, and I've attempted it several times since to little result.  
Let $f$ be an entire function with $|f(z)|\geq 1$ for all $|z|\geq 1$.  Prove that $f$ is a polynomial.
I was trying to use something about $f$ being uniformly convergent to a power series, but I can't get it to go anywhere.

Picard's Theorem proves this instantly; which states:
Let $f$ be a transcendental (non-polynomial) entire function. Then $f-a$ must have infinitely many zeros for every $a$ (except for possibly one exception, called the lacunary value). 
For example, $e^z-a$ will have infinitely many zeros except for $a=0$ and so the lacunary value of $e^z$ is zero. 
Your inequality implies that $f$ and $f-\frac{1}{2}$ have only a finite number of zeros. Thus $f$ cannot be transcendental. 
Of course this is what we call hitting a tac with a sludge hammer. A more realistic approach might be the following: 
Certainly $f$ has a finite number of zeros say $a_1,\ldots,a_n$, so write $f=(z-a_1)\cdots(z-a_n)\cdot h$, where $h$ is some non-zero entire function. Then the inequalities above give us
$|\frac{1}{h}|\le \max\left\{\max_{z\in D(0,2)}|\frac{1}{h}|, |(z-a_1)\cdots (z-a_n)|\right\}$ on the entire complex plane. 
Said more simply $|\frac{1}{h}|<|p(z)|$ for every $z\in C$ for some polynomial $p$. That implies $\frac{1}{h}$ is a polynomial. But remember that $\frac{1}{h}$ is nonzero, so $h$ is a constant and $f$ must therefore be a polynomial. 
