Why is this corollary of Liouville true I know that if $f(z)$ is an entire function and $f(z)$ is bounded, then $f(z)$ is constant. I also know how this can be used to prove the fundamental theorem of algebra.
However, in my books after Liouville it states as a corollary
Suppose that f is entire and n a natural number such that $$|f(z)| \le K(1+|z|^{n})$$ for all  $z \in \mathbb{C}$, then f must be a polynomial of degree at most n.
But I am a bit confused by this, for one, $(1+|z|^{n}) \ge 1$ so wouldn't this simply just imply that f is constant again, by Liouville.
Im sure there is a big flaw in my reasoning somewhere and I am looking to sort it out. Anyways, if this is indeed the case, then what is the proof? I think it probably wouldn't be a very long one seeing as the book just slipped it in at the end of the page as a remark.
Thanks

By Cauchy's Integral formula, we see that
\begin{align}
|f^{(n)}(z)| \leq C \int_{C_R} \frac{|f(\zeta)|}{|z-\zeta|^{n+1}}\ |d\zeta| \leq \int_{C_R} \frac{(1+|\zeta|^n)}{\left||\zeta|-|z| \right|^{n+1}}|d\zeta| \leq C\frac{(1+R^n)R}{|R-|z||^{n+1}} \leq C
\end{align}
for all $z$. Hence $f^{(n)}(z)$ is constant. 
Moreover, it's easy to see $f(z) = z^n+1$ satisfies the bound and it's not constant. 
