Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$. 
Evaluate $\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$

I sense the answer has some connection with $e$, but I don't know how it is. Please help. Thank you.

For $\frac{P(n)}{(n-r)!},$ where $P(n)$ is a polynomial.
If the degree of $P(n)$ is $m>0,$ we can write $P(n)=A_0+A_1(n-r)+A_2(n-r)(n-r-1)+\cdots+A_m(n-r)(n-r-1)\cdots\{(n-r)-(m-1)\}$
Here $k^2=C+B(k-1)+A(k-1)(k-2)$
Putting $k=1$  in the above identity, $C=1$
$k=2,B+C=4\implies B=3$
$k=0\implies 2A-B+C=0,2A=B-C=3-1\implies A=1$
or comparing the coefficients of $k^2,A=1$
So, $$\frac{k^2}{(k-1)!}=\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}=\frac1{(k-3)!}+\frac3{(k-2)!}+\frac1{(k-1)!}$$
$$\sum_{k=1}^\infty \frac{k^2}{(k-1)!}=\sum_{k=1}^\infty  \left(\frac{(k-1)(k-2)+3(k-1)+1}{(k-1)!}\right)$$
$$=\sum_{k=3}^\infty \frac1{(k-3)!}+\sum_{k=2}^\infty \frac3{(k-2)!}+\sum_{k=1}^\infty \frac1{(k-1)!}$$ as $\frac1 {(-r)!}=0$ for $r>0$
$=e+3e+e=5e$
