Set-up
Let $f: \mathbb{N} \to \mathbb{C}$ be a multiplicative arithmetic function (meaning that $f(mn) = f(m)f(n)$ whenever $m$ and $n$ are relatively prime). Suppose that for every $\epsilon >0,$ that $$f(n) \ll_\epsilon n^\epsilon,$$ here I am using Vinogradov’s notation, this means that $|f(n)| \leq C_\epsilon n^\epsilon$ for some constant $C_\epsilon$ which depends on $\epsilon.$.
Suppose further that for each prime $p, f(p)=g $ for some $g \geq 1.$ This $g$ is thus independent of $p.$.
Consider now, for a prime p, the identity of formal power series $$1+gX+f(p^2)X^2 + \cdots = (1-X)^{-g} h_p(X).$$ It is then easy to see that $h_p(0)=1,h_p’(0)=0.$.
Question.
Let now $b_p(k)$ be the coefficient of $X^k$ in $h_p(X).$ I have seen it claimed that the coefficients $b_p(k)$ are bounded by a polynomial in $k$ that is independent of $p.$
Could someone explain why this holds?
Here's my naive approach.
Let $a(x) =1+gX+f(p^2)X^2 + \cdots =\sum_{n-0}^{\infty} a_n x^n $ so that $(1-x)^{-g} h_p(x) =a(x) $.
Then
$\begin{array}\\ h_p(x) &=a(x)(1-x)^g\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{g-j}x^{g-j}\\ &=\sum_{n=0}^{\infty} a_n x^n \sum_{j=0}^g \binom{g}{j}(-1)^{j}x^{j}\\ &=\sum_{n=0}^{\infty} x^n\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}\\ \end{array} $
and
$\begin{array}\\ |\sum_{i=0}^{\min(n, g)} a_{n-i} \binom{g}{i}(-1)^{i}| &\le \sum_{i=0}^{\min(n, g)} |a_{n-i} \binom{g}{i}|\\ &\le \sum_{i=0}^{\min(n, g)} g^if(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} f(p^{n-i})\\ &\le g^g\sum_{i=0}^{\min(n, g)} C_{\epsilon}(n-i)^{\epsilon}\\ &\le g^{g+1}C_{\epsilon}n^{\epsilon}\\ \end{array} $