Proof of Equation (2.5)
\begin{equation*}
h_{k-r}^{c}={\exp(-c)}\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{c}^{i}}}{i!}\left[ \sum\limits_{j=0}^{\infty }{\frac{{{r}^{j}}}{{{\left( k+i+1 \right)}^{j+1}}}} \right] \right\}}.
\end{equation*}
Proof:
Performing long division for ${1}/{\left[ \left( k+1+i \right)-r \right]}$, we derive
\begin{equation*}
\frac{1}{\left( k+1+i \right)-r}=\frac{{{r}^{0}}}{\left( k+1+i \right)}+\frac{{{r}^{1}}}{{{\left( k+1+i
\right)}^{2}}}+\frac{{{r}^{2}}}{{{\left( k+1+i \right)}^{3}}}+\cdots.
\end{equation*}
Therefore,
\begin{align*}
{{c}^{k+1-r}}{\exp(c)}h_{k-r}^{c}&={\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{{{c}^{i}}}{i!\left( k+1+i-r \right)}}\\
&={\exp(-c)}\sum\limits_{i=0}^{\infty }{\left\{ \frac{{{c}^{i}}}{i!}\left[ \sum\limits_{j=0}^{\infty }{\frac{{{r}^{j}}}{{{\left( k+i+1 \right)}^{j+1}}}} \right] \right\}}.
\end{align*}
$\square$