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I'm working through my Combinatorics textbook and am stuck on this proof. The textbook explains it pretty well, but I am having trouble with one of the steps. I was hoping I could get some help here

Theorem: If A(x) and B(x) are formal power series and the constant term of B(x) is zero, then A(B(x)) is a formal power series.

Proof:

Let $A(x) = a_0 + a_1x + a_2x^2 + ....$ and $B(x) = xC(x)$ This is so that B(x) has no constant term

$$A(B(x)) = A(xC(x))$$ $$= a_0 + a_1xC(x) + a_2x^2(C(x))^2 + ...$$

For each $k \gt 0$, $a_kx^kC(x)^k$ is a formal power series since it is the product of formal power series. Also, for each $m \lt k$, $[x^m](a_kx^kC(x))^k = 0$. This is because when $x^k$ is distributed into $C(x)^k$, the smallest degree of x will be k, so all coefficients of powers of x less than k will be 0. This next part is where I'm lost

Therefore $$[x^n]A(B(x)) = [x^n](a_0 + a_1xC(x) + a_2x^2C(x)^2 + ...)$$ $$= [x^n](a_0 + a_1xC(x) + a_2x^2C(x)^2 + ... + a_nx^nC(x)^n)$$

I don't understand how we could have gone from the first line to the second. Could someone please be kind enough to explain this to me? I have a feeling it's simple, but for some reason I just can't see it.

Thanks

The_Questioner
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    Every power of $x$ appearing in $a_kx^kC(x)^k$ is at least $k$ (because of the factor of $x^k$). If $n<k$, then $x^n$ does not appear in this term (or, if you prefer, appears with coefficient $0$). Thus, the only terms that contribute to the $x^n$ term of $A(B(x))$ are those with $k\le n$. – Brian M. Scott Jun 25 '16 at 21:12
  • Thanks. I knew it was simple. I just couldn't see it. – The_Questioner Jun 25 '16 at 21:21
  • @The_Questioner See this answer for a natural notion of convergence for such formal power series that makes clear this and related matters. – Bill Dubuque Jun 25 '16 at 21:45

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