Evaluating the sum: $S_n=\sum_{k=1}^\infty\dfrac{k^n}{(k!)^2}$
By using Stirling numbers of the second kind we have: $$ k^n = \sum_{j=0}^{n}{n \brace j}(k)_j $$ where $(k)_j$ is the falling Pochhammer symbol: $(k)_j = k(k-1)\cdot\ldots\cdot(k-j+1).$ Since: $$\sum_{k=1}^{+\infty}\frac{(k)_j}{(k!)^2}=I_{-j}(2)$$ it follows that:
$$\sum_{k=1}^{+\infty}\frac{k^n}{(k!)^2}=\sum_{j=0}^{n}{n \brace j}I_{-j}(2)=\sum_{j=0}^{n}{n \brace j}I_{j}(2).$$
Have a look at this OEIS entry, too.
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{S}_{n + 2} & = \sum_{k = 1}^{\infty}{k^{n + 2} \over \pars{k!}^{2}} = \sum_{k = 1}^{\infty}{k^{n} \over \bracks{\pars{k - 1}!}^{2}} = \sum_{k = 0}^{\infty}{\pars{k + 1}^{n} \over \pars{k!}^{2}} = 1 + \sum_{k = 1}^{\infty}{1 \over \pars{k!}^{2}} \sum_{\ell = 0}^{n}{n \choose \ell}k^{\ell} \\[3mm] & = 1 + \sum_{\ell = 0}^{n}{n \choose \ell} \sum_{k = 1}^{\infty}{k^{\ell} \over \pars{k}!^{2}} \end{align}
\begin{equation} \imp\quad S_{n + 2} = 1 + \sum_{\ell = 0}^{n}{n \choose \ell}S_{\ell} \,;\qquad n \geq 0\,,\quad \left\lbrace\begin{array}{rcl} \ds{S_{0}} & \ds{=} & \ds{\,\mathrm{I}_{0}\pars{2} - 1} \\[1mm] \ds{S_{1}} & \ds{=} & \ds{\,\mathrm{I}_{1}\pars{2}} \end{array}\right.\tag{1} \end{equation}
$$ \mbox{A few of them,}\quad \left\lbrace\begin{array}{rcrcr} \ds{S_{2}} & \ds{=} & \ds{\mathrm{I}_{0}\pars{2}}&& \\[1mm] \ds{S_{3}} & \ds{=} & \ds{\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{4}} & \ds{=} & \ds{2\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{2\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{5}} & \ds{=} & \ds{5\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{4\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{6}} & \ds{=} & \ds{13\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{10\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{7}} & \ds{=} & \ds{36\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{29\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{8}} & \ds{=} & \ds{109\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{90\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{9}} & \ds{=} & \ds{359\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{295\,\mathrm{I}_{1}\pars{2}} \\[1mm] \ds{S_{10}} & \ds{=} & \ds{1266\,\mathrm{I}_{0}\pars{2}} & \ds{+} & \ds{1030\,\mathrm{I}_{1}\pars{2}} \end{array}\right. $$