Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to 
$$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^3-1)}{n(n-2)!}\right)$$
But this didn't give me any clues to solve this.
I just guess telescoping is the way to solve because this is finite-sum, but I don't have any idea.

Write the sum like
$$\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\frac{(-1)^n n}{n!} + \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=1}^{100}\frac{(-1)^n n}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}+\sum_{n=0}^{99}\frac{(-1)^{n+1}}{n!} +\sum_{n=1}^{100} \frac{(-1)^n}{n!}$$
$$=\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}.$$
Because only the zeroth term of the second sum and the 100th term of the third sum survive.  Now your expression is
$$100! \times \left(1+\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!}-1+\frac{1}{100!}\right)$$
$$= 100!\sum_{n=1}^{100} \frac{(-1)^n n^2}{n!} +1.$$
If $n$ is odd and at least three, then two successive terms of the remaining sum are
$$\frac{-n}{(n-1)!}+\frac{n+1}{n!} = \frac{-1}{(n-2)!}+\frac{1}{n!}.$$  So the sum from 3 to 100 telescopes to $-1 + 1/99!.$  Tacking on the $n=1$ and $n=2$ terms gives us
$$100!\left(\frac{-1}{0!} +\frac{2}{1!} -1 +\frac{1}{99!}\right) +1 = 101.$$
