Proof of Equation (3.15)
\begin{equation*}
\frac{{{\partial }^{\left( n \right)}}\left( h_{s}^{c} \right)}{\partial {{s}^{n}}}=\exp (-c)\sum\limits_{i=0}^{\infty }{\frac{{{\left( -1
\right)}^{n}}n!{{c}^{i}}}{i!{{\left( s+1+i \right)}^{n+1}}}}
\end{equation*}Proof:
\begin{align*}
\frac{\partial \left( h_{s}^{c} \right)}{\partial s}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1
\right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{2}}}}, \\
\frac{{{\partial }^{2}}\left( h_{s}^{c} \right)}{\partial {{s}^{2}}}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1
\right)\left( -2 \right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{3}}}}, \\
& \vdots \\
\frac{{{\partial }^{n}}\left( h_{s}^{c} \right)}{\partial {{s}^{n}}}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1
\right)\cdots \left( -n \right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{n+1}}}}.
\end{align*}
The above formulas conclude the first to $n$th partial derivative of $h_{s}^{c}$ on $s$.
$\square$