Let $f$ be an entire function and suppose there is a constant $k>0$, and $R > 0$, and $n \in \mathbb N$ such that $|f(z)| > k|z|^n$ for $|z| > R$. Show that f is a polynomial and determine its degree.
I am almost sure my proof isn't correct. Hence why I really need help with this
My (Poor) Attempt
$f$ is entire and hence has a derivative at $0$ meaning the function;
and $$F(z) = f'(0)$$ when $z=0$
$$F(z)= \frac{f(z)-f(0)}{z}$$ otherwise.
It is well defined and continuous everywhere.
Since $f$ is entire, it has a power series expansion about $0$ that converges everywhere, which gives an expansion for $F(z)$ that also converges everywhere; therefore, $F$ is entire.
Now we see that if $|f(z)|> k|z|^{n-1}$ for $|z| > R$, then we can take $|z| > R$, which forces
$|F(z)| > \frac{|f(z)|+|f(0)|} {|z|} > k|z|^n + \frac{|f(0)|}{ R}$
Thus, if we instead take $|z| > max(R,1, |f(0)| R )$,
we will have $|F(z)|>(k + 1)|z|^n$. As a result , $ F $ is a polynomial of degree $> n$,
so away from $0$, $f$ is a polynomial of degree at most $n - 1$.
Of course, by continuity, this means $f$’s value at $0$ comes from the polynomial as well, and we are done.
