Entire functions such that $f(z^{2})=f(z)^{2}$

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Characterize the entire functions such that $f(z^{2})=f(z)^{2}$ for all $z\in \mathbb{C}$. Hint: Divide in the cases $f(0)=1$ and $f(0)=0$. For the first case prove (I've alredy done this) and use that $f(z^{2^{n}})= f(z)^{2^{n}}$ for all $n$ natural to see that $f$ is constant. For the second case, if $f$ is not identically zero, then $f$ has a zero in $z=0$ of order $m\geq 1$.

In the second case we can find $m\in\mathbb{N}^*$ such that $f(z) = z^mg(z)$ with $g$ entire and $g(0)\neq 0$. We have $f(z^2)=z^{2m}g(z^2)=z^{2m}g(z)^2$ hence $g(z^2)=g(z)^2$ for all $z\in\mathbb{C}$. Since $g(0)\neq 0$, $g$ is constant (use the first case) and $f$ has the form $f(z)=Cz^m$ with $C\in\mathbb{C}$. If we go back to the equation we should have $C^2z^{2m}=Cz^{2m}$ hence $C^2=C$ and $C=0$ or $C=1$. Finally, the solutions are $f=0$ and $f(z)=z^m$, $m\in\mathbb{N}$.

Let's begin by classifying formal power series $f$ in $\mathbb{C}[[Z]]$ such that $f(Z^2) = f(Z)^2.$

Note that any power of the identity or 0 satisfies the above equation i.e. every element of $\{f:\mathbb{C} \rightarrow \mathbb{C} : f(Z) = Z^n \text{ for some } n\in\mathbb{N}\}\cup\{0\}$ satisfies $f(Z^2) = f(Z)^2$. We demonstrate that that is the complete set of solutions

Let $f\in \mathbb{C}[[Z]]$ and write

$$f(Z) = a_nZ^n + a_kZ^k + ... $$

where $k>n$ and $a_n \neq 0.$

Then

$$f(Z)^2 = a_n^2 Z^{2n} + 2a_n a_k Z^{n+k} \mod Z^{n+k+1}$$

and

$$f(Z^2) = a_n Z^{2n} \mod Z^{n+k+1}.$$

If $f(Z^2) = f(Z)^2$, it follows $a_k = 0$ and $a_n = a_n^2.$ Thus, as $a_n \neq 0,$ we obtain $a_n = 1.$

Hence, $f(Z) = Z^n$ for some $n.$

As the ring of entire functions embeds into the ring of formal power series over $\mathbb{C},$ we conclude the only entire functions commuting with $Z^2$ are powers of the identity or 0.