The proof is a straightforward application of Corollary \[crll:iteratedgammasum\]. There is a constant $C$ such that $g_n^l \leq C \alpha^n \Gamma(n+\beta)$ for all $n\in {\mathbb{N}}_0$ and $l \in [1,L]$. Accordingly, $$\begin{aligned}
 \left| [x^n] \prod_{l=1}^L \left( g^l(x) \right)^{t_l} \right| \leq \alpha^n C^{|\textbf{t}|} \sum_{\substack{\textbf{m} \in {\mathbb{N}}_0^{|\textbf{t}|} \\ |\textbf{m}| = n}} \prod_{r=1}^{|\textbf{t}|} \Gamma(m_{r}+\beta) & & \forall\textbf{t} \in {\mathbb{N}}_0^L \text{ and } n \in {\mathbb{N}}_0. \end{aligned}$$ An application of Corollary \[crll:iteratedgammasum\] results in the lemma.

\[crll:power\_prod\_estimate\] If $g^{1}, \ldots, g^{L} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$ with $g_0^{l}=0$, then there exists a constant $C \in {\mathbb{R}}$ such that $$\begin{aligned}
 \left| [x^n] \prod_{l=1}^L \left( g^l(x) \right)^{t_l} \right| &\leq C^{|\textbf{t}|} \alpha^n \Gamma(n+\beta-|\textbf{t}| +1) && \forall \textbf{t} \in {\mathbb{N}}_0^L, n \in {\mathbb{N}}_0 \text{ with } 1 \leq|\textbf{t}| \leq n. \end{aligned}$$

As a consequence of Proposition \[prop:betashift\], $\frac{g^l(x)}{x} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+1}}$. Hence, $$\begin{aligned}
 \left| [x^{n-|\textbf{t}|}] \prod_{l=1}^L \left( \frac{g^l(x)}{x} \right)^{t_l} \right| \leq C^{|\textbf{t}|} \alpha^{n-|\textbf{t}|} \Gamma(n-|\textbf{t}|+\beta+1) && \forall \textbf{t} \in {\mathbb{N}}_0^L, n \in {\mathbb{N}}_0 \text{ with } 1 \leq |\textbf{t}| \leq n. \end{aligned}$$

The composition of two power series can be expressed as the sum $$\begin{gathered}
 h(x) = \sum_{\substack{\textbf{t} \in {\mathbb{N}}_0^{L}}} f_{t_1,\ldots,t_L} \prod_{l=1}^L \left(g^{l}(x)\right)^{t_l}, \\ \intertext{which can be split in preparation for the extraction of asymptotics:} = \sum_{\substack{\textbf{t} \in {\mathbb{N}}_0^{L} \\ |\textbf{t}| \leq R}} f_{t_1,\ldots,t_L} 