Sum of the series $\sum _{ r=1 }^{ \infty }{ \frac { { r }^{ 3 } }{ r! } } $ How do I prove that the sum of the series 
$1+\frac { { 2 }^{ 3 } }{ 2! } +\frac { { 3 }^{ 3 } }{ 3! } +\frac { { 4 }^{ 3 } }{ 4! } +.... = 5e$
 A: Note that $r^3=r((r-1)(r-2)+3(r-1)+1)$. Hence
$$\sum _{ r=1 }^{ \infty }\frac {r^{ 3 } }{ r! }=
\sum _{ r=1 }^{ \infty }\frac {r(r-1)(r-2) }{ r! }
+3\sum _{ r=1 }^{ \infty }\frac {r(r-1)}{ r! }+\sum _{ r=1 }^{ \infty }\frac {r}{ r! }.$$
Can you take it from here? Show your efforts!
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{r = 1}^{\infty}{r^{3} \over r!} & =
\left.\pars{x\,\totald{}{x}}^{3}\sum_{r = 1}^{\infty}{x^{r} \over r!}
\right\vert_{\ x\ =\ 1} =
\left.\pars{x\,\totald{}{x}}^{3}\pars{\expo{x} - 1}
\right\vert_{\ x\ =\ 1} = \bbx{5\expo{}}
\end{align}
A: hint
$$\sum_{n=1}^\infty \frac {n^3}{n!}=$$
$$\sum_{n=1}\frac {n^2-1+1}{(n-1)!}=$$
$$\sum_{n=2} \frac {n-2+3}{(n-2)!}+e=$$
$$\sum_{n=3} \frac {1}{(n-3)!}+3e+e=$$
$$5e $$
