Proving that two sequences of polynomials defined over partitions are inverse to each other

Question

For any fixed $c>0$ consider the polynomials \begin{align*} & p_n(X_1,X_2,\ldots) := \frac{n!}{c} \sum\limits_{b=1}^n \frac{c^b}{b!(n+1-b)!} \sum\limits_{\substack{l_1,\ldots,l_b \geq 1 \\ l_1+\cdots+l_b=n}} X_{l_1} \cdots X_{l_b} \ , \tag{1} \end{align*} which we can also write as a sum over partitions as \begin{align*} & p_n(X_\bullet) = \frac{n!}{c} \sum\limits_{\pi \vdash n} \frac{\operatorname{sym}(\pi) c^{\ell(\pi)}}{\ell(\pi)!(n+1-\ell(\pi))!} X_{\pi_1} \cdots X_{\pi_{\ell(\pi)}} \ , \tag{2} \end{align*} where $\ell(\pi)$ is the length of the partition and $\operatorname{sym}(\pi)$ denotes the number of distinct ways in which the numbers $(\pi_1,\ldots,\pi_{\ell(\pi)})$ can be arranged. I was looking for polynomials $q_n$ with the 'inversion'-property \begin{align*} & X_n = q_n(p_1(X_\bullet),p_2(X_\bullet),\ldots) \tag{3} \end{align*} and with some help from the computer and OEIS have found the candidate \begin{align*} q_n(Y_1,Y_2,\ldots) = & -\frac{1}{c} \sum\limits_{a=1}^n \frac{(-c)^a (n+a-2)!}{a!(n-1)!} \sum\limits_{\substack{k_1,\ldots,k_a \geq 1 \\ k_1+\cdots+k_a=n}} Y_{k_1} \cdots Y_{k_a} \tag{4}\\ = & -\frac{1}{c} \sum\limits_{\tau \vdash n} \frac{(-c)^{\ell(\tau)} \operatorname{sym}(\tau) (n+\ell(\tau)-2)!}{\ell(\tau)! (n-1)!} Y_{\tau_1} \cdots Y_{\tau_{\ell(\tau)}} \ . \tag{5} \end{align*} With the computer I have checked the property (3) up to $n=25$, where I have used formulas (2) and (5) for the polynomials. Unfortunately, proving this seems to require methods for partitions that I have little experience with.

Any help is much appreciated!

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Edit: No idea if this helps, but as $q_n(p_1(X_\bullet),p_2(X_\bullet),\ldots)$ has the form $\sum\limits_{\eta \vdash n} c^{\ell(\eta)-1} C_\eta X_{\eta_1} \cdots X_{\eta_{\ell(\eta)}}$ for some coefficients $C_\eta$ and one can relatively easily see $C_{(n)} = 1$, it suffices to show that the map \begin{align*} & c \mapsto q_n(p_1(X_\bullet),p_2(X_\bullet),\ldots) \end{align*} is constant.

Also here are the first five polynomials: \begin{align*} & p_1(X) = X_1\\ & p_2(X) = X_2 + c X_1^2\\ & p_3(X) = X_3 + 3c X_2X_1 + c^2 X_1^3\\ & p_4(X) = X_4 + 4c X_3X_1 + 2c X_2^2 + 6c^2 X_2X_1^2 + c^3 X_1^4\\ & p_5(X) = X_5 + 5c X_4X_1 + 5c X_3X_2 + 10c^2 X_3X_1^2 + 10c^2 X_2^2X_1 + 10c^3 X_2 X_1^3 + c^4 X_1^5 \end{align*} \begin{align*} & q_1(Y) = Y_1\\ & q_2(Y) = Y_2 - c Y_1^2\\ & q_3(Y) = Y_3 - 3c Y_2Y_1 + 2c^2 Y_1^3\\ & q_4(Y) = Y_4 - 4c Y_3Y_1 - 2c Y_2^2 + 10c^2 Y_2Y_1^2 - 5c^3 Y_1^4\\ & q_5(Y) = Y_5 - 5c Y_4Y_1 - 5c Y_3Y_2 + 15c^2 Y_3Y_1^2 + 15c^2 Y_2^2Y_1 - 35c^3 Y_2Y_1^3 + 14c^4 Y_1^5 \end{align*}

Answer 1

These are the inverse pair OEIs A134264 and OEIS A350499.