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.

Let $D_M=\{z:|f(z)|<M\}$ open connected set; since $f$ is non-constant $D_M$ is not the plane for any $M$; given any Jordan curve $J \subset D_M$, the interior of $J$ is contained in $D_M$ by maximum modulus, hence $D_M$ is simply connected.
Let now $B_{2r}$ a small disc of radius $2r$ around $0$ where $f$ vanishes only at $z=0$ and $2a=\min_{|z|=r}|f(z)| >0$; since then $f(\partial B_r) \cap D_a = \emptyset$ it follows that $D_a \subset B_r$ (as otherwise if there is $|w|>r, |f(z)| <a$ there is a path in $D_a$ joining $w$ and $0 \in B_r$ and that must intersect $\partial B_r$) so in particular $D_a$ bounded and $f$ vanishes only at zero.
It follows that $f$ is a polynomial since $\infty$ is not an essential singularity ($|f(z)| > a, |z|>r$ and since $f$ vanishes only at $0$ we are done!
