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I am currently going through the proof of the Theorem $5.4.2$ on page 38 of Enumerative Combinatorics Vol $2$ by R. Stanley and I think I understand the proof. The proof of Corollary $5.4.3$ as shown in the attached proof is just a sketch and I am unable to prove it using the hint given. All my reasoning does not lead me anywhere. Could some please show me the proof? or perhaps a hint? I saw some proofs that strictly used the Cauchy-Residue Theorem but I want to avoid using this.

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    $\begingroup$ I have to say that this looks pretty clear to me. As Stanley says, for $H(x)=x^k$ this reduces to the previous theorem. In general if $H(x)=\sum h_k x^k$, then for a given $n$, the $x^n$ coefficient in $H(f(x))$ will depend only on finitely many of the $h_k$, so one can assume $H(x)$ is a polynomial etc. $\endgroup$ – Lord Shark the Unknown Oct 18 '18 at 3:47
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A little elaboration on Lord Shark the Unknown's comment. Fix $n$ and $f$ and define an operator $T: K[[x]] \rightarrow K$ by

$$T(H) = n[x^n]H(f(x)) - [x^{n-1}]H'(x) G(x)^n.$$

We want to show $T(H) = 0$ for any $H$. Note that $T(H) = 0$ when $H(x) = x^k$ for any $k$. This is exactly the statement of Theorem 5.4.2. Note also that this operator is linear. You can show directly that $T(\alpha H_1 + \beta H_2) = \alpha T(H_1) + \beta T(H_2)$.

Thus $\ker T$ contains every polynomial. Now for any $H$, let $H_n$ be the truncation of $H$ to an $n$-th degree polynomial. Then $T(H) = T(H_n)$, since the $x^{n+1}$ or higher terms do not affect the coefficient of $x^n$ in $H(f(x))$ or the coefficient of $x^{n-1}$ in $H'(x) G(x)^n$. Since $H_n$ is a polynomial, it follows that $T(H_n) = 0$, and so $T(H) = 0$.

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