Let $f$ be an entire function such for some $N \in \mathbb{N}$ and $R >0$, the following property holds:
$|f(z)| \geq |z|^N$ $\forall z \in \mathbb{C}$ with $|z| \geq R$.
Show that $f$ is a polynomial of degree greater than or equal to $N$.
Progress thus far:
Clearly, $f$ is not identically zero.
Case 1) If $f$ is also nowhere 0, it's not hard to show $f$ is bounded and therefore, by Louisville's Theorem, a constant.
Case 2) If $f$ is neither identically zero nor nowhere 0, then we can show that it has a finite number of zeros in the ball $B(0,R)$. From these zeros we can get a polynomial $p$, whose zeros are exactly those of $f$, and an entire function $g$ such that $f = pg$. Consequently, $g$ is nowhere zero.
At this point, I'd like to show $g \equiv c$, where $c$ is some complex constant.
A more general problem has been solved here, but I'm trying to avoid using big hammers like Casorati-Weierstrass. (When this problem was assigned, we hadn't covered it yet.)
Best Answer
You're almost there, and your idea in case 2 is the right one. Though we can argue more simply by just using $z^N$ instead of introduction a new polynomial $p$.
Consider the function $g(z)=\frac{z^N}{f(z)}$. This function is holomorphic except possibly at the points where $f(z)$ equals zero. But we know that $|g(z)| \leq 1$, so each of these points is actually a removable singularity of $g(z)$ so $g$ extends to a bounded holomorphic function, and is thus constant.
It's interesting to note here that there was nothing special about $|f(z)| \geq |z|^N$. The inequality could have been the other way and the result would still hold. More generally, no entire function can dominate another entire function without being a constant multiple of one another.