Bell me differentiate me!

Relevant wiki: Bell Numbers

Similar solution with @Naren Bhandari's

$\begin{aligned} S & = \sum_{\color{#3D99F6}n=2}^\infty \frac {n^7+n^6+n^5+n^4+n^3+n^2+n+1}{(n-2)!} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {(n+2)^7+(n+2)^6+(n+2)^5+(n+2)^4+(n+2)^3+(n+2)^2+n+2+1}{n!} \\ & = \sum_{n=0}^\infty \frac {n^7 + 15n^6 + 97n^5 + 351n^4 + 769n^3 + 1023n^2 + 769n + 255}{n!} & \small \color{#3D99F6} \because \text{ Bell number }B_n = \frac 1e \sum_{k=0}^\infty \frac {k^n}{k!} \\ & = B_7e + 15B_6e + 97B_5e + 351B_4e + 769B_3e + 1023B_2e + 769B_1e + 255B_0e \\ & = \big(877 + 15(203) + 97(52) + 351(15) + 769(5) + 1023(2) + 769(1) + 255(1)\big)e \\ & = 21146 e \end{aligned}$

Therefore, $A = \boxed{21146}$ .

Method 1 Given $\sum_{n=2}^{\infty} \left(\dfrac{n^7+n^6+n^5+\cdots+n^2+n+1}{(n-2)!}\right)$ Now multiplying and dividing by $n(n-1)$ we can deduce that $\dfrac{n^7+n^6+n^5+\cdots+n^2 +n+1}{(n-2)!} =\dfrac{n^9}{n!} -\dfrac{n}{n!}$ Hence our sum become $S = \sum_{n=1}^{\infty} \left(\dfrac{n^9}{n!} -\dfrac{n}{n!} \right) = B_9 e - B_1e = 21146e$ where $B_9,B_1$ are ninth and first Bells numbers.

Note: We have that $\sum_{j=0}^{\infty}\dfrac{x^j}{j!}= B_n e\implies B_0 = \dfrac{1}{e}\sum_{j=0}^{\infty}=1$ where $B_n$ is the nth Bells numbers where $n\in\mathbb Z\geq 0$ . Further the Bells numbers are generated by $B_{n+1} = \sum_{k=0}^{n} B_k \binom{n}{k}\implies B_1 = \sum_{k=0}^{0}B_0\binom{0}{0} =1$ Set $n= 1 ,2, 3,\cdots ,8$ $\begin{aligned} B_{2} & = \sum_{k=0}^{1} B_k \binom{1}{k}=1+1 =2 \\ B_{3} & = \sum_{k=0}^{2} B_k \binom{2}{k}= 1+3+1 = 5\\ B_{4} & = \sum_{k=0}^{3} B_k \binom{3}{k} = 1+6+7+1 =15 \\ B_{5} & = \sum_{k=0}^{4} B_k \binom{4}{k} = 1+10+25+15+1=52 \\ B_{6} & = \sum_{k=0}^{5} B_k \binom{3}{k} = 1+15+65+90+31+1=203 \\ B_{7} & = \sum_{k=0}^{6} B_k \binom{3}{k} =1+21+140+350+301+63+1= 877 \\ B_{8} & = \sum_{k=0}^{7} B_k \binom{7} {k} = 1+28+266+1050+1701+966+127+1=41470 \\ B_{9} & = \sum_{k=0}^{8} B_k \binom{8}{k} = 1+36+462+2646+6951+7770+3025+255+1=21147\end{aligned}$

Method 2

Given $\sum_{n=2}^{\infty} \left(\dfrac{n^7+n^6+n^5+\cdots+n^2+n+1}{(n-2)!}\right)$ Now multiplying and dividing by $n(n-1)$ we can deduce that $\dfrac{n^7+n^6+n^5+\cdots+n^2 +n+1}{(n-2)!} =\dfrac{n^9}{n!} -\dfrac{n}{n}$ Hence our sum become $S = \sum_{n=1}^{\infty} \left(\dfrac{n^9}{n!} -\dfrac{n}{n!} \right) =\sum_{n=1}^{\infty} \dfrac{n^9}{n!} - e =21147e-e =21146e$ To determine the $\sum_{n=1}^{\infty} \dfrac{n^9}{n!}=221147e$ we will be using following technique .Since $\sum_{k=0}^{\infty}\dfrac{x^k}{k!} =e^x$ Now we perform differentiation with respect to x and multiply by $x$ after each differentiation as $\begin{aligned} \sum_{k=0}^{\infty}\dfrac{k\cdot x^{k-1}}{k!} & =e^x\implies \sum_{k=0}^{\infty}\dfrac{k\cdot x^{k}}{k!} = xe^x \\ \sum_{k=0}^{\infty}\dfrac{k^2\cdot x^{k-1}}{k!} & =( x+ 1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^3\cdot x^{k-1}}{k!} &=( x^2+3x+ 1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^4\cdot x^{k-1}}{k!} & =(x^3 + 6x^2 +7x + 1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^5\cdot x^{k-1}}{k!} &= (x^4+ 10x^3+25x^2+15x+1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^6\cdot x^{k-1}}{k!} & = (x^5+15x^4+65x^3+ 90x^2+ 31x+ 1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^7\cdot x^{k-1}}{k!} &=( x^6+21x^5+ 140x^4+ 350x^3+ 301x^2+ 63x+ 1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^8\cdot x^{k-1}}{k!}& =(x^7 + 28x^6+266x^5 +1050x^4+1701x^3+966x^2+127x+1)e^x\\ \sum_{k=0}^{\infty}\dfrac{k^9\cdot x^{k-1}}{k!} &=(x^8+36x^7+462x^6+2646x^5+6951x^4+7770x^3+3025x^2+255x+1)e^x\end{aligned}$ Now setting $x=1$ we obtain $e, 2e , 5e, 15e, 52e,203e, 877e, 4140e , 21147e$ .