General quotient rule for a couple of special functions problem formulation
Given a discrete probability density $\{p(i)\}_{i=0}^\infty$, let $G(z)$ be the following power series (called probability generating function)
\begin{equation}G(z)\triangleq \sum_{i=0}^\infty p(i)\cdot z^i\end{equation}
and abbreviate as $G^{(n)}(z)$ the $n$-th derivative of $G(z)$ with respect its argument $z\in\mathbb{R}$,
\begin{equation}G^{(n)}(z)\triangleq \frac{\text{d}^{n}\,G(z)}{\text{d}z^n}\end{equation}
I have to resolve the following problem: compute the $n$-th derivative, where $n$ is an arbitrary integer, of the ratio $G^{(1)}(z)/G(z)$, i.e. I have to find the explicit expression of
\begin{equation}\zeta_n(z)\triangleq \left(\frac{G^{(1)}(z)}{G(z)}\right)^{(n)}\end{equation}
in function of the given density $p(i)$.

partial solution
a quick search (e.g. see here) tells me that a possible approach consist into consider $G^{(1)}/G$ as $G^{(1)}\cdot 1/G$ and then to apply the Leibnitz rule, so
\begin{equation}\begin{aligned}
\zeta_n(z)&=\sum_{k=0}^n\binom{n}{k}\cdot\left(G^{(1)}(z)\right)^{(k)}\cdot\left(\frac{1}{G(z)}\right)^{(n-k)}\\
&=\sum_{k=0}^n\binom{n}{k}\cdot G^{(1+k)}(z)\cdot\left(\frac{1}{G(z)}\right)^{(n-k)}\\
\end{aligned}\end{equation}
here the first derivative $G^{(1+k)}(z)$ is straightforward because $G(z)$ is a power series, in fact for any integer $k$
\begin{equation}\begin{aligned}
G^{(k)}(z)&\triangleq\frac{\text{d}^k}{\text{d}z^k}\left[\sum_{i=0}^\infty p(i)\cdot z^i\right]=
\sum_{i=0}^\infty p(i)\cdot\frac{\text{d}^k}{\text{d}z^k}\left[z^i\right]=
\sum_{i=0}^\infty p(i)\cdot \left(P_k^i\cdot z^{i-k}\right)\\
\end{aligned}\end{equation}
where I have introduced the permutation coefficient
\begin{equation}P_k^i\triangleq \begin{cases}\frac{i!}{(i-k)!} & \text{if } i \geq k \\ 0 & \text{otherwise}\end{cases}\end{equation}
the derivative $G^{(k)}(z)$ is still a power series, indeed by exploiting the fact that $P_k^i=0$ when $i<k$,
\begin{equation}\begin{aligned}
G^{(k)}(z)&=\sum_{i=k}^\infty p(i)\cdot \left(P_k^i\cdot z^{i-k}\right)=\sum_{i=0}^\infty p(i+k)\cdot \left(P_k^{i+k}\cdot z^{i}\right)\\
&=\sum_{i=0}^\infty p(i+k)\cdot \left((i+k)!\cdot z^{i}\right)=\sum_{i=0}^\infty p_k(i)\cdot z^{i}
\end{aligned}\end{equation}
where I have introduced the non-negative sequence (I have a strong suspect, but I'm not sure, that in general it is not a discrete density anymore - however this is an off topic problem)
\begin{equation}p_k(i)\triangleq (i+k)!\cdot p(i+k)\end{equation}

question
the real problem is to compute for any $k$ the derivative
\begin{equation}\left(\frac{1}{G(z)}\right)^{(k)}\tag{1}\end{equation}
this derivative can be computed by using the Faa' di Bruno's formula, which if I'm not wrong is a chain rule of arbitrary order, because $1/G(z)$ can be written as the composition
\begin{equation}\frac{1}{x}\bigg|_{x=G(z)}\end{equation}
however, before to dig into the rabbit hole of complex computations, I would like to have some external suggestion about how to compute $(1)$.
My question is the following:
it is really necessary to use Faa' di Bruno or, due to the special form of $G(z)$ (which I know it is not a generic function but a nice power series), it is possible to use some clever method?
 A: I think that a better answer to this question is the general derivative formula
\begin{equation}\label{Sitnik-Bourbaki-reform}\tag{1}
\frac{\textrm{d}^n}{\textrm{d}x^n}\biggl[\frac{u(t)}{v(t)}\biggr]
=(-1)^n\frac{|W_{(n+1)\times(n+1)}(t)|}{v^{n+1}(t)},
\end{equation}
where $u(t)$ and $v(t)\ne0$ are differentiable functions, $U_{(n+1)\times1}(t)$ is an $(n+1)\times1$ matrix whose elements $u_{k,1}(t)=u^{(k-1)}(t)$ for $1\le k\le n+1$, $V_{(n+1)\times n}(t)$ is an $(n+1)\times n$ matrix whose elements
\begin{equation*}
v_{i,j}(t)=
\begin{cases}
\dbinom{i-1}{j-1}v^{(i-j)}(t), & i-j\ge0\\
0, & i-j<0
\end{cases}
\end{equation*}
for $1\le i\le n+1$ and $1\le j\le n$, and $|W_{(n+1)\times(n+1)}(t)|$ denotes the determinant of the $(n+1)\times(n+1)$ lower Hessenberg matrix
\begin{equation*}
W_{(n+1)\times(n+1)}(t)=\begin{bmatrix}U_{(n+1)\times1}(t) & V_{(n+1)\times n}(t)\end{bmatrix}.
\end{equation*}
For more information on this formula \eqref{Sitnik-Bourbaki-reform}, please refer to https://math.stackexchange.com/a/4261705/945479.
