How do I show that if $f$ is entire and $\{\lvert f(z)\rvert < M\}$ is connected for all $M$, then $f$ is a power function? 
Let $f$ be a non constant entire function satisfying the following conditions :

*

*$f(0)=0$

*for every positive real $M$, the set $\{z: \left|f(z)\right|<M\}$ is connected.

Prove that $f(x)=cz^n$ for some constant $c$ and positive integer $n$.

Let $f(z)=a_nz^n+\cdots+a_1z+a_0$ be function that satisfies the given conditions.   As $f(0)=0$ we have $a_0=0$ and $f(z)=a_nz^n+\cdots+a_1z$.
As $f$ is non-constant function, its zeros are isolated. So, there exists an $r>0$ such that $f$ is non-zero on $B_r=\{z:|z|<r\}$. I was thinking of connecting this to connectedness of $\{z: \left|f(z)\right|<M\}$.
I wanted to check what goes wrong in case of $f(z)=z^2+z$.  I want to check if the given set is connected for this but failed in doing so.

Suppose $f$ has Taylor series
$$
f(z)=\sum_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum_{n=0}^{\infty}\frac{a_n}{z^n}
$$
If $f$ is not polynomial, then $\infty$ is an essential singularity of $f$ ($0$ is an essential singularity of $g$). By Casorati-Weierstrass theorem, for $0\in \Bbb{C}$, there is a sequence $z_n'\to0$ such that $\lim_{n\to\infty}g(z_n')=0$, i.e. there is $z_n=1/z_n'\to\infty$ such that $\lim_{n\to\infty}f(z_n)=0$. This means that $f$ has always $z_0$ that $f(z_0)=0$ near $\infty$.
Since zero of analytic function is isolated and $f(0)=0$, by choosing $M$ so that $z_0\in \{z: \left|f(z)\right|<M\}$ $(0\in \{z: \left|f(z)\right|<M\}$ is obvious$)$, we conclude that $\{z: \left|f(z)\right|<M\}$ can not be connected because $z_0$ and $0$ are separated. Thus $f$ must be polynomial. 
If $f$ has nonzero root, i.e. $f(z_1)=0,\: z_1\ne 0$, then $\{z: \left|f(z)\right|<M\}$ including $z_1$ can not be connected for $z_1$ and $0$ are separated. So $f(z)=0$ can only has zero root. Thus we conclude that $f(z)=cz^n$.
