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.

We may assume that $$
f(z)=cz^n(1+c_1z+c_2z^2+\cdots)=cz^ng(z),
$$
where $c>0$ (for simplicity) and $g(z)$ has no zeros in $|z|\le r_0$ for some positive $r_0$.
Furthermore there is a positive $r_1<r_0$  such that 
\begin{align}
|f(z)|>\frac{c{r_1}^n}{2}  \quad \text{on}\,\;|z|=r_1,\tag{1}
\end{align}
since $|g(z)|\to 1$ $(\,z\to 0\,)$.
Let $M=\displaystyle\frac{c{r_1}^n}{2}$ and $A=\{z:\, |f(z)<M\}$.
If $g(z_0)=0$ for some point $z_0\,(\,|z_0|>r_0\,)$, then $A$ contains $0$ and $z_0$, and hence by the connectedness there is a curve $\gamma \subset A$ joining $0$ and $z_0$.  But this is impossible by $(1)$.
Therefore $g(z)\ne 0$ and hence we can write $g(z)=e^{-h(z)}$ , $h(z)=\sum_{k=1}^\infty a_kz^k$.  
Suppose that there exists $a_k\ne 0$. Let $H(r)=\max_{|z|=r} \operatorname{Re}\, h(z)$. 
Then as well-known
$$
|a_k|r^k\le \max\{ 4H(r), 0\}-2\operatorname{Re}\, h(0)= 4H(r).$$
Hence we have 
\begin{align}
\min_{|z|=r} |g(z)|&=\min_{|z|=r} \left|e^{-h(z)}\right|=e^{-H(r)}\le e^{-|a_k|r^k/4}.
\end{align}
Thus for sufficiently large $r$ we have 
$$
|f(z_0)|\le cr^n\cdot e^{-|a_k|r^k/4}<M$$
at a point $z_0$ on $|z|=r$.
Then $A$ contains $z_0$, but it is impossible.
Thus all $a_k=0$ and we have $f(z)=cz^n$.
