Entire Function Problem in Complex Analysis I am currently working on some review problems in complex analysis and came upon the following conundrum of a problem.
"If $f(z)$ is an entire function, and satisfies $|f(z^2)|\le|f(z)|^2$, prove that f(z) is a polynomial."
My intuition tells me to show that f(z) has a pole at infinity by showing that infinity is not an essential or removable singularity. However, I am getting stuck after this.
Thanks for the help, 

Let
$$
M=\sup_{|z|=2}|f(z)|
$$
Then, with the condition given, it can be proven inductively that
$$
\sup_{|z|=2^{2^n}}|f(z)|\le M^{2^n}
$$
which implies
$$
|f(z)|\le|z|^{2\log_2(M)}
$$
We can use Cauchy's Theorem to give
$$
f^{(n)}(z)=n!\int_{\gamma_R}\frac{f(w)\,\mathrm{d}w}{(w-z)^{n+1}}
$$
where $\gamma_R$ is the circle of radius $R$. If we choose $n\gt2\log_2(M)$, then if we let $R\to\infty$,
$$
\begin{align}
|f^{(n)}(z)|
&\le\int_{\gamma_R}\frac{|w|^{2\log_2(M)}\,\mathrm{d}w}{(z-w)^{n+1}}\\
&\sim2\pi R^{2\log_2(M)-n}\\[12pt]
&\to0
\end{align}
$$
Since the $n^\text{th}$ derivative is identically $0$, $f$ must be a polynomial.
