polynomial in entire function with polynomial coefficients Let $f$ be an entire function. Assume that there are polynomials $p_0, \ldots , p_n(z)$, not all zero, such that $p_n(z)(f(z))^n + p_{n−1}(z)(f(z))^{n−1} + \cdots + p_0(z) = 0$. Prove that f is a polynomial.
 A: We may assume $p_n(z)\ne 0.$ Suppose $f$ is not a polynomial. Then
$$
p_0(z)=-f(z)\left[p_n(z)(f(z))^{n-1} + p_{n-1}(z)(f(z))^{n-2} + \cdots + p_1(z)\right]$$
yields 
$$
p_n(z)(f(z))^{n-1} + p_{n-1}(z)(f(z))^{n-2} + \cdots + p_1(z)=0,$$
since $p_0(z)$ is a polynomial. Then
$$p_1(z)=-f(z)\left[p_n(z)(f(z))^{n-2} + p_{n-1}(z)(f(z))^{n-3} + \cdots + p_2(z)\right]$$
yields$$
p_n(z)(f(z))^{n-2} + p_{n-1}(z)(f(z))^{n-3} + \cdots + p_2(z)=0.$$
Eventually we have
$$p_n(z)f(z) + p_{n-1}(z)=0,$$
and $$p_{n-1}(z)=-p_n(z)f(z)$$ yields $p_n(z)=0$. It's a contradiction.
A: We can assume $p_n\not \equiv 0.$ The given equality implies
$$\tag 1 |p_n(z)||f(z)|^n = |\sum_{k=0}^{n-1}p_{k}(z)f(z)^{k}| \le \sum_{k=0}^{n-1}|p_{k}(z)||f(z)|^{k}.$$
We'll show $(1)$ implies $|f(z)|$ has no more than polynomial growth at $\infty.$ As is well known, this implies $f$ is a polynomial.
Let $E= \{z\in \mathbb C\: |f(z)|\le 1\}.$ Obviously $|f(z)|$ has no more than polynomial growth as $z\to \infty$ within $E.$
Now consider $z\in \mathbb C \setminus E.$ Because $|f(z)|>1,$ we can say that by $(1),$
$$ |p_n(z)||f(z)|^n   \le |f(z)|^{n-1}
\sum_{k=0}^{n-1}|p_{k}(z)|.$$
Dividing by $|f(z)|^{n-1}$ then gives
$$|p_n(z)||f(z)| \le \sum_{k=0}^{n-1}|p_{k}(z)|.$$
Because $p_n \not \equiv 0,$ there is a constant $c>0$ such that $|p_n(z)|>c$ for large $|z|.$ For such $z\in \mathbb C \setminus E,$ we then have
$$c|f(z)| \le \sum_{k=0}^{n-1}|p_{k}(z)|.$$
This implies $|f(z)|$ has has no more than polynomial growth as $z\to \infty$ within $\mathbb C \setminus E,$ and the proof is finished.
