<s>[INST] Compute $\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}$ Compute
$$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!}.$$

Today I have participated in Olympiad which had 8 different tasks. This was one of them.  
I've started by substituting concrete values and looking for a pattern:
$\sum_{0}^{\infty}(-1)^k \frac{(k + 1) ^ 2}{k!} = 1 - \frac{2^2}{1!} + \frac{3^2}{2!} - \frac{4^2}{3!} + \frac{5^2}{4!} + \dots$
This gave me insight that it may be connected to the following series:
$e = 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$
After that for a few hours I was trying to transform original series to the series for $e$. My intuition tells me that the result is something similar to $e \times constant$. But I wasn't able to prove it or disprove it.
 [/INST] HINT:
As $(k+1)^2=k(k-1)+3k+1,$
For $k\ge2,$
$$\dfrac{(k+1)^2}{k!}=\dfrac1{(k-2)!}+3\cdot\dfrac1{(k-1)!}+\dfrac1{k!}$$
$$\dfrac{(k+1)^2}{k!}x^k=x^2\cdot\dfrac{x^{k-2}}{(k-2)!}+3x\cdot\dfrac{x^{k-1}}{(k-1)!}+\dfrac{x^k}{k!}$$
For $k=0,$  $$\dfrac{(k+1)^2}{k!}x^k=1$$
For $k=1,$  $$\dfrac{(k+1)^2}{k!}x^k=4x=3x\cdot\dfrac1{0!}=\dfrac{x^1}{1!}$$
Now use $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
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