+ n_{d-1}}}{n_{1}! \ldots n_{d-1}!} \exp (\beta e^{\nu_{d-k} b^k C_{d-k-1,d-1}}  + \nu_{d-k} b^k C_{d-k,d-1} ) - \\
\nonumber \sum_{n_{1}=0}^{p} \ldots \sum_{n_{d-1}=0}^{p} \frac{\beta^{n_{1}+\ldots + n_{d-1}}}{n_{1}! \ldots n_{d-1}!} \exp (\beta e^{\nu_{d-k} b^k p}  + \nu_{d-k} b^k C_{d-k,d-1} ). \end{gathered}$$ In the last step $(p+1)^{d-1}$ terms of infinite series were changed. Then we estimate the exponential part of series from above by $\exp(\beta e^{\nu_{d-k}p})$.\
Each term of both finite series tends to zero when divided by $e^{\beta(d-k-1)}$ (as $\beta$ tends to infinity), so that the whole series tends to zero. This step is repeated till we get $$\begin{gathered}
\nonumber \alpha_{p} \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-k}}}{n_{1}! \ldots n_{d-k}!} \exp ( \nu_{d-k} b^k C_{d-k,d-k} ) \leq \\
\nonumber \alpha_{p} \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-k}}}{n_{1}! \ldots n_{d-k}!} \exp \left( \nu_{d-k} b^k  \left[ \sum_{i=1}^{d-k-1}n_{i}n_{d-k}+ \sum_{i=1}^{d-k}n_{i} \right] \right) = \\
\nonumber \alpha_{p} \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k-1}=0}^{\infty} \frac{(\beta e^{\nu_{d-k} b^k})^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! \ldots n_{d-k-1}!} \exp (\beta e^{\nu_{d-k} b^k (\sum_{i=1}^{d-k-1}n_{i}+1)}) \leq \end{gathered}$$ $$\begin{gathered}
\nonumber \alpha_{p} \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k-1}=0}^{\infty} \frac{(\beta e^{\nu_{d-k} b^k})^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! \ldots n_{d-k-1}!} \exp (\beta e^{\nu_{d-k} b^k q})+ \\
\nonumber \alpha_{p} \sum_{n_{1}=0}^{q} \ldots \sum_{n_{d-k-1}=0}^{q} \frac{(\beta e^{\nu_{d-k} b^k})^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! \ldots n_{d-k-1}!} \exp (\beta e^{\nu_{d-k} b^k (\sum_{i=1}^{d-k-1}n_{i}+1)}) - \\
\nonumber \alpha_{p} \sum_{n_{1}=0}^{q} \ldots \sum_{n_{d-k-1}=0}^{q} \frac{(\beta e^{\nu_{d-k} b^k})^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! 