Sum to infinity

$\begin{aligned} &e^x = 1+ \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\cdots \\ &\text{Differentiating w.r.t x on both sides}\\ &e^x = \frac{1}{1!} + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \cdots\cdots \\ &\text{Multiplying both sides by x }\\ &xe^x = \frac{x}{1!} + \frac{2x^2}{2!} + \frac{3x^3}{3!} + \frac{4x^4}{4!} + \cdots\cdots \\ &\text{Again differentiating w.r.t x on both sides}\\ &e^x(x+1) = \frac{1}{1!} + \frac{2^2x}{2!} + \frac{3^2x^2}{3!} + \frac{4^2x^3}{4!} + \cdots\cdots \\ &\text{Again Multiplying both sides by x }\\ &x(x+1)e^x = \frac{x}{1!} + \frac{2^2x^2}{2!} + \frac{3^2x^3}{3!} + \frac{4^2x^4}{4!} + \cdots\cdots \\ &\text{Again differentiating w.r.t x on both sides}\\ &(x^2 + 3x +1)e^x = \frac{1}{1!} + \frac{2^3x}{2!} + \frac{3^3x^2}{3!} + \frac{4^3x^3}{4!} + \cdots\cdots \\ &\text{Put x=1 , we get}\\ & 5e = \large 1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\cdots \\ \end{aligned}$

$\begin{aligned} S & = 1 + \frac {2^3}{2!} + \frac {3^3}{3!} + \frac {4^3}{4!} + \cdots \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^3}{n!} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^3}{(n+1)!} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+1)^2}{n!} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {n^2+2n+1}{n!} \\ & = \underbrace{\frac 1{0!}}_{n=0} + \underbrace{\frac 4{1!}}_{n=1} + \sum_{\color{#3D99F6}n=2}^\infty \frac {n(n-1)+3n+1}{n!} \\ & = 5 + \sum_{\color{#20A900}n=2}^\infty \left(\frac 1{(n-2)!} + \frac 3{(n-1)!} + \frac 1{n!} \right) \\ & = 5 + \sum_{\color{#D61F06}n=0}^\infty \frac 1{n!} + 3 \sum_{\color{#3D99F6}n=1}^\infty \frac 1{n!} + \sum_{\color{#20A900}n=2}^\infty \frac 1{n!} \\ & = 5 + \sum_{\color{#D61F06}n=0}^\infty \frac 1{n!} + 3 \left(\sum_{\color{#D61F06}n=0}^\infty \frac 1{n!} - \frac 1{0!} \right) + \sum_{\color{#D61F06}n=0}^\infty \frac 1{n!} - \frac 1{0!} - \frac 1{1!} \\ & = 5 \sum_{\color{#D61F06}n=0}^\infty \frac 1{n!} = \boxed{5e} \end{aligned}$

$\displaystyle\sum_{n=1}^{\infty}\frac {n^3}{n!} = \displaystyle\sum_{n=1}^{\infty}\frac {n^2-1+1}{(n-1)!} =\displaystyle\sum_{n=1}^{\infty}\left (\frac {n+1}{(n-2)!} + \frac {1}{(n-1)!}\right)$

$=\displaystyle\sum_{n=1}^{\infty}\left (\frac{1}{(n-3)!} + \frac {3}{(n-2)!} + \frac {1}{(n-1)!}\right) = 5e$

Hence $k=5$ .

Bell numbers