Let's do some calculus! (41)

Credits for the approach applied to my solution go to Anubhav Tyagi .

---

Using the series expansion of $e^x$ ,

$e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \ ... \$

We differentiate both sides w.r.t. $x$ once to obtain,

$e^x = \dfrac{1}{1!} + \dfrac{2x}{2!} + \dfrac{3x^2}{3!} + \dfrac{4x^3}{4!} + \ ... \$

Multiplying by $x$ on both sides of the equation yields,

$xe^x = \dfrac{x}{1!} + \dfrac{2x^2}{2!} + \dfrac{3x^3}{3!} + \dfrac{4x^4}{4!} + \ ... \$

Differentiate again to get,

$e^x (x+1) = \dfrac{1}{1!} + \dfrac{2^2 x}{2!} + \dfrac{3^2 x^2}{3!} + \dfrac{4^2 x^3}{4!} + \ ... \$

We observe that the following can be written as,

$e^x (x+1) = \displaystyle \sum_{n=1}^{\infty} x^{n-1} \dfrac{n^2}{n!}$

From here, we've obtained a sum where the exponent of $n$ in the numerator is $2$ . So, by similar approach, if we continue to differentiate w.r.t. $x$ and multiply by $x$ alternatively, we'll obtain a sum where the exponent of $n$ in the numerator is $8$ .

After some tedious work , we obtain,

$e^x (x^7 + 28x^6 + 266x^5 + 1050x^4 + 1701x^3 + 966x^2 + 127x + 1) = \dfrac{1}{1!} + \dfrac{2^8 x}{2!} + \dfrac{3^8 x^2}{3!} + \dfrac{4^8 x^3}{4!} + \ ... \ = \displaystyle \sum_{n=1}^{\infty} x^{n-1} \dfrac{n^8}{n!}$

Putting $x = -1$ , we obtain,

$-\dfrac{50}{e} = \dfrac{1}{1!} - \dfrac{2^8}{2!} + \dfrac{3^8}{3!} - \dfrac{4^8}{4!} + \ ... \ = \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \dfrac{n^8}{n!}$

Thus,

$k = -50 \implies |k| = \boxed{50}$

---

The part (tedious work) skipped form the main solution.

$e^x (x+1) = \dfrac{1}{1!} + \dfrac{2^2 x}{2!} + \dfrac{3^2 x^2}{3!} + \dfrac{4^2 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x+1) = \dfrac{x}{1!} + \dfrac{2^2 x^2}{2!} + \dfrac{3^2 x^3}{3!} + \dfrac{4^2 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^2+3x+1) = \dfrac{1}{1!} + \dfrac{2^3 x}{2!} + \dfrac{3^3 x^2}{3!} + \dfrac{4^3 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x^2+3x+1) = \dfrac{x}{1!} + \dfrac{2^3 x^2}{2!} + \dfrac{3^3 x^3}{3!} + \dfrac{4^3 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^3+6x^2+7x+1) = \dfrac{1}{1!} + \dfrac{2^4 x}{2!} + \dfrac{3^4 x^2}{3!} + \dfrac{4^4 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x^3+6x^2+7x+1) = \dfrac{x}{1!} + \dfrac{2^4 x^2}{2!} + \dfrac{3^4 x^3}{3!} + \dfrac{4^4 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^4+10x^3+25x^2+15x+1) = \dfrac{1}{1!} + \dfrac{2^5 x}{2!} + \dfrac{3^5 x^2}{3!} + \dfrac{4^5 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x^4+10x^3+25x^2+15x+1) = \dfrac{x}{1!} + \dfrac{2^5 x^2}{2!} + \dfrac{3^5 x^3}{3!} + \dfrac{4^5 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^5+15x^4+65x^3+90x^2+31x+1) = \dfrac{1}{1!} + \dfrac{2^6 x}{2!} + \dfrac{3^6 x^2}{3!} + \dfrac{4^6 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x^5+15x^4+65x^3+90x^2+31x+1) = \dfrac{x}{1!} + \dfrac{2^6 x^2}{2!} + \dfrac{3^6 x^3}{3!} + \dfrac{4^6 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^6+21x^5+140x^4+350x^3+301x^2+63x+1) = \dfrac{1}{1!} + \dfrac{2^7 x}{2!} + \dfrac{3^7 x^2}{3!} + \dfrac{4^7 x^3}{4!} + \ ... \$

Multiplying both sides by $x$ ,

$e^x x(x^6+21x^5+140x^4+350x^3+301x^2+63x+1) = \dfrac{x}{1!} + \dfrac{2^7 x^2}{2!} + \dfrac{3^7 x^3}{3!} + \dfrac{4^7 x^4}{4!} + \ ... \$

Differentiating both sides w.r.t. $x$ ,

$e^x (x^7 + 28x^6 + 266x^5 + 1050x^4 + 1701x^3 + 966x^2 + 127x + 1) = \dfrac{1}{1!} + \dfrac{2^8 x}{2!} + \dfrac{3^8 x^2}{3!} + \dfrac{4^8 x^3}{4!} + \ ... \$