If $f,g$ are entire functions such that $f(g(z))=0, \forall z, $ then $g$ is constant or $f(z) =0, \forall z \ ?$ Let $f,g$ be entire functions such that $f(g(z))=0, \forall z.$ Could anyone advise me on how to prove/disprove: either $g(z)$ is constant or $f(z) =0, \forall z \ ?$
Hints will suffice, thank you.

Another way to make it: Assume that $g$ is not constant in particular $g'\neq 0$: 
As $f\circ g=0$, then by derivation we get $g'\times (f'\circ g)=0$, but the ring of entires functions ( $\Bbb C$ connected) is an integral domain, it follow that $f'\circ g=0$, another derivation we get $g'\times (f''\circ g)=0$, hence $f''\circ g=0$, by same argument we can show that for all $n\in \Bbb N$; $f^{(n)}\circ g=0$.
Now let $a\in \Bbb C$ and $b=g(a)$, since $f$ is analytic we have :
$$\forall z\in \Bbb C\ \ \ f(z)=\sum_{n=0}^{+\infty}\frac{f^{n}(b)}{n!}(z-b)^n=0$$
because $f^{(n)}(b)=f^{(n)}(g(a))=0$
