Shifted character products: exact for $p_1^{\mathrm{\#}}$ and leading terms for $p_k^{\mathrm{\#}}$

Exact product with $p_1^{\mathrm{\#}}$: fix $\lambda ⊢n$ with $n\ge \mathrm{\mid }\sigma \mathrm{\mid }+1$ (for $n<\mathrm{\mid }\sigma \mathrm{\mid }$ both sides vanish; at $n=\mathrm{\mid }\sigma \mathrm{\mid }$ one has $p_{\sigma \cup 1}^{\mathrm{\#}}=0$ and $p_1^{\mathrm{\#}}=\mathrm{\mid }\sigma \mathrm{\mid }$, so the identity holds directly). By [F1], $p_{\sigma }^{\mathrm{\#}}(\lambda )p_1^{\mathrm{\#}}(\lambda )=n^{\downarrow r}{\chi }_{\sigma \cup 1^{n-r}}^{\lambda }\mathrm{/}\dim \lambda \cdot n$ and $p_{\sigma \cup 1}^{\mathrm{\#}}(\lambda )=n^{\downarrow (r+1)}{\chi }_{\sigma \cup 1^{n-r}}^{\lambda }\mathrm{/}\dim \lambda$, so the claim reduces to $n^{\downarrow r}\cdot n=n^{\downarrow (r+1)}+n^{\downarrow r}r$, which is the defining recursion $n^{\downarrow (r+1)}=n^{\downarrow r}(n-r)$ rewritten; hence $p_{\sigma }^{\mathrm{\#}}{p}_1^{\mathrm{\#}}=p_{\sigma \cup 1}^{\mathrm{\#}}+\mathrm{\mid }\sigma \mathrm{\mid }{p}_{\sigma }^{\mathrm{\#}}$.

Equality-case analysis: let $\rho$ be such that $f_{\sigma (k)}^{\rho }\mathrm{\ne }0$ and ${\mathrm{deg}}_1(p_{\rho }^{\mathrm{\#}})={\mathrm{deg}}_1(p_{\sigma }^{\mathrm{\#}})+k$, where $\tau =(k)$ and ${\mathrm{deg}}_1(p_k^{\mathrm{\#}})=k$ because $m_1((k))=0$ for $k\ge 2$. By [F2] the equality case forces the overlap $X_1\cap X_2$ to consist of common nontrivial cycles of $s_1$ and $s_2^{-1}$; since $s_2$ is a single $k$-cycle, either $X_1\cap X_2=\mathrm{\varnothing }$, giving $\rho =\sigma \cup k$ with coefficient $f_{\sigma (k)}^{\sigma \cup k}=1$, or $X_1\cap X_2$ is one common $k$-cycle, which requires $m_k(\sigma )\ge 1$, gives $\rho =(\sigma \setminus k)\cup 1^k$, and forces $X_2\subseteq X_1$.

Coefficient in the second case: in the case $\rho =(\sigma \setminus k)\cup 1^k$ abbreviate $m:=m_k(\sigma )\ge 1$ and $\mathrm{\ell }:=m_1(\sigma )$; a direct computation from $z_{\mu }={\prod }_i{i}^{m_i(\mu )}{m}_i(\mu )!$ gives $\frac{z_{\rho }}{z_{\sigma }{z}_{(k)}}=\frac{(k+\mathrm{\ell })!}{\mathrm{\ell }!k^{2}m}$, so by [F2] the identity $f_{\sigma (k)}^{\rho }=km$ is equivalent to $g_{\sigma (k)}^{\rho }=\frac{(k+\mathrm{\ell })!}{\mathrm{\ell }!k}$. The partial-permutation count recorded in [F2] in this case is the number of ways to choose a $k$-cycle inside the $(k+\mathrm{\ell })$-point fixed-point set of $w_{\rho }$: all other cycles of $w_{\rho }$ must remain unchanged: choose the $k$-point support, $\left(\genfrac{}{}{0px}{}{k+\mathrm{\ell }}{k}\right)=\frac{(k+\mathrm{\ell })!}{k!\mathrm{\ell }!}$ ways, and a $k$-cycle on it, $(k-1)!$ ways, giving $g_{\sigma (k)}^{\rho }=\frac{(k+\mathrm{\ell })!}{\mathrm{\ell }!k}$ as required; hence $f_{\sigma (k)}^{\rho }=k{m}_k(\sigma )$, the factor $m=m_k(\sigma )$ counting the choice of which $k$-part of $\sigma$ is the common cycle.

Conclusion: every other contributing $\rho$ has ${\mathrm{deg}}_1(p_{\rho }^{\mathrm{\#}})<{\mathrm{deg}}_1(p_{\sigma }^{\mathrm{\#}})+k$ by the definition of the equality case in step 1.2, so the expansion takes the displayed form; specialising $\sigma =(k^m)$ gives $m_k(\sigma )=m$ and $(\sigma \setminus k)\cup 1^k=(k^{m-1},1^k)$, which is the stated recurrence. No choice principle is used.