Challenging identity regarding Bell polynomials

Question

Note: [2015-03-08] A proof of the identity below was aimed to close the gap of a rather extensive elaboration of this answer of mine. The identity (1) below is part of a more complex one, which is stated in Part 3, (39)-(42) in this follow-up answer. The split in two answers was necessary due to the restriction of up to $30000$ characters per answer. I could prove the other parts, but this final part still needed a verification.

In the meanwhile I found an answer and the solution is now provided as part of my follow-up answer starting there with expression (48).

I've checked the validity of the identity for small values of $n$. It was also checked in somewhat different representations all along the work of my related answer.

In order to keep the calculations manageable, I've introduced some abbreviations:

Let's consider a function $f=f(z)$ and its Taylor series expansion at a point $x$ \begin{align*} f(z+x)=\sum_{n\geq 0}\frac{f^{(n)}(x)}{n!}z^n \end{align*} then we use the following abbreviation for the Bell polynomials \begin{align*} B^{f}_{n,k}(x):=B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)}) \end{align*} We use the Pochhammer symbol \begin{align*} \left(f(x)\right)_k:= f(x)f(x-1)\cdot\ldots\cdot f(x-k+1) \end{align*} and a relationship with the Stirling numbers of the first kind, namely \begin{align*} (x)_n=\sum_{k=0}^{n}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}x^{k} \end{align*}

I have also omitted the argument $x$ in the identity below, so for example

\begin{align*} \frac{(\ln\circ g)^{m-j}}{g^k} := \frac{\left(\ln(g(x))\right)^{m-j}}{\left(g(x)\right)^k} \end{align*}

Problem: Show that the identity is valid for $n \geq 2, 1\leq l \leq n-1$ and $1\leq m \leq l$:

\begin{align*} \sum_{k=1}^{n-l}&\sum_{j=0}^{m}\binom{m}{j} \frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{n-l,k}^{g}\\ &=\sum_{j=m}^{n}\sum_{q=1}^{j}\sum_{k=q}^{n-l}(-1)^{k-q}\frac{q!}{(j-m)!}\\ &\qquad\qquad\cdot\binom{m}{j-q}\begin{bmatrix}k\\q\end{bmatrix}f^{j-m} \frac{ (\ln \circ g)^{j-q}}{g^k}B^{g}_{n-l,k} \end{align*}

Please note, that the indices of the Bell polynomials of the LHS and RHS in (1) coincide.

Answer 1

This is not an answer but just presenting current work i have done on the problem

We know that:

$$ (f)_n = \sum_{k=0}^n (-1)^{n-k}{n\brack k} f^k $$

Therefore

$$ \frac{d^j}{df^j}[(f)_k] = \sum_{v=j}^k (-1)^{k-v}{k \brack v} \frac{v!}{(v-j)!} f^{v-j} $$

Therefore by placing this in the formula presented, (Using $x_{n-l,k}$ for $B_{n-l,k}^g$)

$$ \sum_{k=1}^{n-l} \sum_{j=0}^m {m \choose j} \frac{d^j}{df^j}[(f)_k] \frac{\ln(g)^{m-j}}{g^k} X_{n-l,k} $$

Then turns into

$$ \sum_{k=1}^{n-l} \sum_{j=0}^m \sum_{v=j}^k {m \choose j} (-1)^{k-v}{k \brack v} \frac{v!}{(v-j)!} f^{v-j} \frac{\ln(g)^{m-j}}{g^k} X_{n-l,k} $$

When looking at the RHS of the expression i see that the $\frac{\ln(g)^{m-j}}{g^k}$ was converted to $\frac{\ln(g)^{p-q}}{g^k}$ Therefore to me it seems that $m=p$ and $j=q$ because i see no reason that this would not be true, But yet i see that the RHS of the expression contains $m$ variables which confuses me, Or perhaps these $m$ variables differ from the LHS to the RHS. Either way i will continue to convert the variables in this way:

$$ \sum_{k=1}^{n-l} \sum_{q=0}^p \sum_{v=q}^k {p \choose q} \frac{v! (-1)^{k-v}}{(v-q)!} {k \brack v} f^{v-q} \frac{\ln(g)^{p-q}}{g^k} X_{n-l,k} $$ Also it interests me that the $q$ variable is in the stirling numbers on the RHS of equation (1), perhaps there is some manipulation with the summations that can be done?

I will add more if needed to this answer when the information presents itself.