Which entire functions satisfy $\,\lvert\,f(z)\rvert \leq \lvert z\rvert^k$? Which entire holomorphic functions satisfy $\,\lvert\,f(z)\rvert \leq \lvert z\rvert^k$, for all $z\in\mathbb{C}$?
So I've shown that $\,\lvert\, f(z)\rvert \leq \lvert z\rvert ^k \implies f(z)$ is a polynomial of degree at most $k$.
Therefore we may write
$f(z)=\sum_{0}^{k} c_nz^n$ where $c_n$ is Taylorian coefficients of $f$.
So I see that this should certainly mean that $c_0=0$ since $\,\lvert\,f(0)\rvert=0$. 
But after that, I have no idea how to proceed. I guess that $c_k \leq 1$ but have no idea to prove it, not to mention how other coefficients would vary according to value of $c_k$.
Any helps are appreciated!
 A: Put $z=\frac{1}{N}$, where $N \in \mathbb{N}$. Then we have:
$$\left|\sum_{n=0}^{k}\frac{c_{n}}{N^n}\right|\leq \frac{1}{N^{k}}$$
Multiply both sides by $N^k$:
$$\left|\sum_{n=0}^{k}c_{n}N^{k-n}\right|\leq 1$$
By triangle inequality:
$$|c_k|-\sum_{n=0}^{k-1}N^{k-n}|c_n|\leq \left|\sum_{n=1}^{k}c_{n}N^{k-n}\right|$$
And inequality:
$$|c_k|-\sum_{n=0}^{k-1}N^{k-n}|c_n|\leq 1$$
Can by satisfy if only if $c_0=c_1=\cdots=c_{k-1}=0$.
A: We have that
$$
\left|\frac{f(z)}{z^k}\right|\le 1, \,\,\,\,
\text{whenever $\,\,z\in\mathbb C\smallsetminus\{0\}$},
$$
and hence the function $g(z)=z^{-k}f(z)$, is holomorphic in 
$\mathbb C\setminus\{0\}$ and 
has removable singularity at $z=0$. Therefore it can be extended holomorphically at $z=0$; in fact it extends to a bounded entire function. Thus $g(z)$ is constant, i.e., $g(z)=c$, for some constant $\,\lvert c\rvert\le 1$.
Therefore, $\,f(z)=cz^k$, with $\lvert c\rvert\le 1 $. 
Note. Let's assume instead that $\lvert\,f(z)\rvert\le \lvert z\rvert^a$, for $a>0$, which is not an integer, then $f\equiv 0$. 
Say $a=n+b$, where $n=\lfloor a\rfloor$ and $b\in (0,1)$. 
Then $g(z)=z^{-n}f(z)$ has a removable singularity at $z=0$, and thus $g$ extends to an entire function, satisfying $\lvert g(z)\rvert\le \lvert z\rvert^b$, for all $z$. But, as $g$ is continuous at $z=0$, then $g(0)=0$, and hence $g(z)=zh(z)$, for some entire function $h$, and hence $\lvert h(z)\rvert\le \lvert z\rvert^{b-1}$. Thus $h$ is bounded, and therefore constant, which implies that $f(z)=cz^{n+1}$, for some $c$. But $\lvert cz^{n+1}\rvert\le \lvert z\rvert^{n+b}$, for all $z$, implies that $c=0$. 
