Let $f$ be an entire function such that the inverse image of the closed unit ball is bounded. State whether $\infty$ is a removable singularity or pole or essential singularity of $f$?
(I am familiar with the following two facts: (1) $f$ has removable singularity at $\infty$ if and only if $f$ is constant. (2) $f$ has a pole at infinity if and only if $f$ is polynomial.)
If $f$ is a constant function $c$ with $|c|>1$ then clearly $f$ has a removable singularity at $\infty$.
Now suppose $f$ is not constant. Then $f$ must have at most finietly many zeroes. But only this fact can't prove that $f$ is a polynomial. (For example $f(z)=ze^z$).
So how to show that $f$ is a polynomial?
The comment by Pythagorous uses a property of essential singularities. If you are not allowed to use that result you can use the following hint:
There exists $R$ such that $|z|>R$ implies $|f(z)| >1$. Let $z_1,z_2,...,z_n$ be the zeros of $f$ (counted according to mutiplicities) in $|z| \leq R$. Let $g=\frac {(z-z_1)(z-z_2)...(z-z_n)} f$. The $g$ is entire and $|g(z)| \leq C|z|^{n}$ for $|z|>R$, for some constant $C$. By a well know result (whose proof you can find on MSE) it follows that $g$ is a polynomial of degree at most $n$. Can you finish?