The numerators diverge much faster this time

It is well known that

$e = \sum_{k=1}^\infty \dfrac{1}{(k-1)!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$

This value is also known (see Calvin's problem)

$\sum_{k=1}^{\infty} \dfrac{ k} {(k-1)!} = \dfrac{1}{0!} + \dfrac{2}{ 1!} + \dfrac{3}{2!} + \dfrac{ 4}{3!} + \cdots \quad = x$ ($x$ is kept secret here so Calvin's problem isn't spoiled.)

So what is

$\sum_{k=1}^{\infty} \dfrac{k^2} {(k-1)!} = \dfrac{1}{0!} + \dfrac{4}{ 1!} + \dfrac{9}{2!} + \dfrac{16}{3!} + \cdots \quad ?$

$$ Notation: $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$. And $0!=1$ as always.