Suppose $f$ is an entire function and that $\exists k &gt; 0, R &gt; 0$ and $n \in \mathbb{N}$ such that $|f(z)| &gt; k|z|^n$ whenever $|z| &gt; R$. Prove that $f$ is a polynomial.  I am not sure how to go about this and I would greatly appreciate some hints.
As in the title, I must prove that: If a complex function $f$ is Entire, and there exist $k,R &gt; 0$, $n \in N$ such that $|f(z)| &gt; k|z|^n$ for all $|z| &gt; R$: Then f is a polynomial.  We've already had a similar question with the inequality reversed, and found that it had to be a polynomial with degree less than n. I was able to show that if it was a polynomial, then it had to have degree less than n, but not that it had to be a polynomial in the first place.  Once again, I'm face with the same problem. I assume this time, the degree of the polynomial will have degree greater than or equal to n, but how do I show that such $f$ must be a polynomial?  It's not like I can go through every other possible function there is, can I? I can show it doesn't work for exponentials, for example, but I just can't think of how to show that this ONLY works for a polynomial...
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