Septaphobia

Please see this note for a simpler problem if you have not already seen it. The method used to solve here is the same as that in the note but much lengthier, as a system of $8$ equations is required, rather than $3.$ The full solution is still posted below.

$W = \frac{1^7}{1!} + \frac{1^7 + 2^7}{2!} + ... \\ = \sum_{n=1}^\infty [\frac{\sum_{k=1}^n k^7}{n!}] \\ = \sum_{n=1}^\infty [\frac{\frac{1}{24}n^2(n+1)^2(3n^4 + 6n^3 - n^2 - 4n + 2)}{n!}] \\ = \frac{1}{24}[\sum_{n=1}^\infty \frac{n(n+1)^2(3n^4 + 6n^3 - n^2 - 4n + 2)}{(n-1)!}]$ Let $S = 24\times W,$ then $S = \sum_{n=1}^\infty \frac{n(n+1)^2(3n^4 + 6n^3 - n^2 - 4n + 2)}{(n-1)!} \\ = \sum_{n=0}^\infty \frac{(n+1)(n+2)^2(3n^4 + 18n^3 + 35n^2 + 24n + 6)}{n!}$ Now define $f(x) = \sum_{n=0}^\infty \frac{(n+1)(n+2)^2(3n^4 + 18n^3 + 35n^2 + 24n + 6)}{n!}x^n$ which is the Taylor series of $g(x) = e^x(3x^7 + 96x^6 + 1064x^5 + 5040x^4 + 10206x^3 + 7728x^2 + 1524x + 24).$ Substitute $x = 1$ and the result is $S = e(3 + 96 + 1064 + 5040 + 10206 + 7728 + 1524 + 24) \\ = 25685e$ and $\boxed{\frac{S}{e} = 25685}.$

Proof that the Taylor series about $x=0$ of $g(x)$ is $f(x).$
The numerator of the terms of $f(x)$ is a polynomial of degree $7,$ and on account of $\frac{x^n}{n!},$ the function whose Maclaurin series is $f(x)$ is of the form $g(x) = e^x(a_1x^7 + a_2x^6 + a_3x^5 + ... + a_7x + a_8).$ The general Maclaurin series of that form is $a_8 + (a_7 + a_8)x + (a_6 + a_7 + \frac{a_8}{2})x^2 + O(x^3)$ Compare this to the first several values of $\frac{(n+1)(n+2)^2(3n^4 + 18n^3 + 35n^2 + 24n + 6)}{n!}x^n$ which are $24,\ 1548x,\ 9264x^2,\ ...$ $a_8 = 24 \\ a_7 = 1548 - 24 = 1524 \\ a_6 = 9264 - 1524 - 12 = 7728 \\ a_5 = 18700 - 7728 - 762 - 4 = 10206 \\ ...$ continuing this method, $a_4 = 5040,\ a_3 = 1064,\ a_2 = 96,\ a_1 = 3.$

Like Caleb's solution, and from equation 27 and apply properties of Bell numbers . The series becomes

$\begin{aligned} & & \frac 1 {24} \displaystyle \sum_{n=1}^\infty \frac {3n^8 + 12n^7 + 14n^6 - 7n^4 + 2n^2}{n!} \\ &= & \frac {e}{24} \left [ 3B_8 + 12B_7 + 14B_6 - 7B_4 + 2B_2 \right ] \\ \end{aligned}$

Substitute those Bell numbers .

Hence the answer is simply $3B_8 + 12B_7 + 14B_6 - 7B_4 + 2B_2 = \boxed{25685}$