The problem statement: Assume $f$ is an entire function and that there is an $n \in \mathbb{N}$ and a $C < \infty$ such that for $z \in C$ $$|f(z)| \le C ( 1+|z|^n)$$ Also assume that $f$ is never zero and $f(0) = 1$. What is the function?
I believe that the problem can be solved in part by using what we refer to as the Cauchy derivative estimates (inequalities in the book). Which tells us:
$$|f^{(n)}(z_0)| \le \frac{n!}{R^n} \, \sup_{|z-z_0| = R} |(f(z)|$$ (where the $|z-z_0| = R$ is allowsing $z$ to be on the boundary of the circle of radius $R$. So by letting $z_0$ be arbitrary, we can use the n-th derivative here and the bound for $f(z)$ to show that for large $R$, this limits to 0.
My question really revolves around whether or not we can say more. This argument tells us that $f$ is a polynomial of at most degree $n$, and as $f(0) = 1$, we know that the constant term is 1, but can we say anything else to answer "What is the function"?
A polynomial without a zero has to be constant.