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How do I go about proving that the composition of formal power series is associative?

I've tried proving the result directly, but the resulting expressions are quite unwieldy. Currently, I'm trying to make use of the topology on $\mathbb{C}[[x]]$, but I can't quite get it to work.

More precisely, I want to prove that if $g(0)=0$ and $h(0)=0$, then $f\circ(g\circ h)=(f\circ g)\circ h$, where $f\circ g(x) :=\sum_{n=0}^\infty f_n (g(x))^n.$

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    @HagenvonEitzen: I edited the question to replace $\neq0$ with $=0$2012-09-11

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A formal power series $g(x)$ such that $g(0)=0$ gives you a continuous ring homomorphism $g^*:K[[x]]\to K[[x]]$ uniquely defined by $x\mapsto g(x)$. In fact, $g^*(f)=f\circ g$ (it is clear if $f$ is a polynomial and extends by continuity to power series). As $h^*(g^*(x))=h^*(g(x))=g(h(x))=(g\circ h)^*(x)$, we have $h^*\circ g^*=(g\circ h)^*$. Evaluating both sides on $f$ we get $(f\circ g)\circ h=f\circ (g\circ h)$.

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    Also see http://planetmath.org/encyclopedia/IAdicTopology.html or http://en.wikipedia.org/wiki/Completion_(ring_theory)#Krull_topology for the topology $K[[x]]$ comes equipped with (any topology that $K$ may have is irrelevant to this topology; it is present for arbitrary coefficient rings).2012-09-11