May I Borrow Your Telescope?

$\begin{aligned} S & = \sum_{k=1}^\infty \frac {(-1)^k(k^2+k+1)}{k!} \\ & = \sum_{k=1}^\infty (-1)^k \left(\frac {k^2}{k!} + \frac {k+1}{k!} \right) \\ & = \sum_{k=1}^\infty (-1)^k \left(\frac {k}{(k-1)!} + \frac {k+1}{k!} \right) \\ & = - \frac 1{0!} - \frac 2{1!} + \frac 2{1!} + \frac 3{2!} - \frac 3{2!} - \frac 4{3!} + \frac 4{3!} + \frac 5{4!} - ... \\ & = - \frac 1{0!} {\color{#3D99F6}- \cancel{\frac 2{1!}} + \cancel{\frac 2{1!}}} {\color{#D61F06} + \cancel{\frac 3{2!}} - \cancel{\frac 3{2!}}} {\color{#3D99F6}- \cancel{\frac 4{3!}} + \cancel{\frac 4{3!}}} {\color{#D61F06} + \cancel{\frac 5{4!}} -} ... \\ & = \boxed{-1} \end{aligned}$

This below is not perfectly my solution and this is Nihar's solution

$\begin{aligned} \Large \sum_{i=1}^{\infty} \frac{(-1)^i (i^2 + i + 1)}{i!} & \Large = \color{#D61F06}{ \sum_{i=1}^{\infty} \frac{(-1)^i i^2}{i!}} \color{#333333}{+} \color{#3D99F6}{ \sum_{i=1}^{\infty} \frac{(-1)^i i}{i!}} \color{#333333}{+} \color{#CEBB00}{ \sum_{i=1}^{\infty} \frac{(-1)^i}{i!}}\\ &\Large = \color{#D61F06}{\sum_{i=1}^{\infty}{\left((-1)^i \frac{(i-1)+1}{(i-1)!}\right)}} \color{#333333}{+} \color{#3D99F6}{{\left(-1+1-\frac{1}{2}-\frac{1}{6}-…\right)}} \color{#333333}{+} \color{#CEBB00}{{\left(-1+\frac{1}{2}-\frac{1}{6}+…\right)}}\\ &\Large =\color{#D61F06}{{\left(\sum_{i=1}^{\infty} {\left( (-1)^i \frac{1}{(i-2)!}\right)}+ \sum_{i=1}^{\infty}{\left((-1)^i \frac{1}{(i-1)!}\right)}\right)}}\color{#333333}{+} \color{#20A900}{(-1)}\\ &\Large =\color{#D61F06}{{\left({\left(0+1-1+\frac{1}{2}-\frac{1}{6}+…\right)}+{\left(-1+1-\frac{1}{2}+\frac{1}{6}\right)}\right)}} \color{#333333}{+}\color{#20A900}{(-1)}\\ &\Large= \color{#624F41}{\boxed{-1}} \end{aligned}$