here uses $\sum_{a = 1}^p \frac{1}{\kappa -a+1} \leq \log\big(\frac{\kappa}{\kappa-p}\big)$ and $C_3 \geq 0$. Equation follows from the fact that for any $x>0$, $\log(1+x) \leq x$. To prove , we have the first order partial derivative of $\P(\theta)$ given by $$\begin{aligned}
 \label{eq:cr2}
\nabla_i \P(\theta) &=&  \I_{\{\Omega^{-1}(i) \leq p \}}\P(\theta)  -  \sum_{\sigma \in \Omega} \Bigg(\frac{\exp\big(\sum_{m = 1}^{p} \theta_{\sigma(m)} \big)}{\prod_{a=1}^{p} \Big(\sum_{m'=a}^{\kappa} \exp\big(\theta_{\sigma(m')}\big) \Big)} \Bigg( \sum_{a = 1}^p \frac{\I_{\{\sigma^{-1}(i) \geq a \}}\exp(\theta_i)}{\sum_{m'=a}^{\kappa} \exp\big(\theta_{\sigma(m')}\big)} \Bigg)   \Bigg) \,.\end{aligned}$$ Define constants $A_1$, $A_2$ and $A_3$ such that $$\begin{aligned}
A_1 & \equiv & \P(\theta) \big|_{\{\theta = {\boldsymbol{0}}\}} = \frac{(p-1)!}{\kappa(\kappa-1)\cdots(\kappa-p+1)}, \label{eq:crA1}\\
A_2 & \equiv & \Bigg( \sum_{a = 1}^p \frac{\exp(\theta_i)}{\sum_{m'=a}^{\kappa} \exp\big(\theta_{\sigma(m')}\big)} \Bigg)\Bigg|_{\{\theta = {\boldsymbol{0}}\}} = \Bigg(\frac{1}{\kappa} + \frac{1}{\kappa-1}+\cdots + \frac{1}{\kappa-p+1}\Bigg), \label{eq:crA2} \\
A_3 & \equiv & \Bigg(\frac{(p-1)(p-2)!}{(p-1)!(\kappa)}  + \frac{(p-2)(p-2)!}{(p-1)!(\kappa-1)} + \cdots + \frac{(p-2)!}{(p-1)!(\kappa-p+2)} \Bigg) \,. \label{eq:crA3}\end{aligned}$$ Observe that, for all $i \in [d]$, $$\begin{aligned}
 \label{eq:cr4}
\nabla_i \P(\theta) \big|_{\{\theta = {\boldsymbol{0}}\}}  = A_1 \Big( \I_{\{\Omega_j^{-1}(i) = p\}}(1 - A_2) + \I_{\{\Omega_j^{-1}(i) < p\}}(1 - A_3) - \I_{\{\Omega_j^{-1}(i) > p\}}A_2 \Big) \;\; \,.\end{aligned}$$ Further define constants $B_1$, $B_2$, $B_3$ and $B_4$ such that $$\begin{aligned}
B_1 &\equiv & \Bigg(\frac{1}{\kappa^2} + \frac{1}{(\kappa-1)^2} + \cdots + \frac{1}{(\kappa - p+1)^2}\Bigg), \label{eq:crB1}\\
B_2 & \equiv & \Bigg(\frac{p-1}{(p-1)\kappa^2}  + \frac{p-2}{(p-1)(\kappa-1)^2} + \cdots + \frac{1}{(p-1)(\kappa-p+2)^2}  \Bigg), \label{eq:crB2} \\
B_3 & \equiv & \Bigg( \frac{(p-1)(p-2)(p-3)!}{(p-1)!\