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.
 A: The most elementary calculation is probably this one:
$$\begin{align*}
\sum_{k\ge 0}\frac{(k+1)^2}{k!}&=1+\sum_{k\ge 1}\frac{k^2+2k+1}{k!}\\
&=\color{red}{1}+\sum_{k\ge 1}\frac{k}{(k-1)!}+2\sum_{k\ge 1}\frac1{(k-1)!}+\color{red}{\sum_{k\ge 1}\frac1{k!}}\\
&=\sum_{k\ge 1}\frac{k-1+1}{(k-1)!}+2\sum_{k\ge 0}\frac1{k!}+\color{red}{\sum_{k\ge 0}\frac1{k!}}\\
&=\color{blue}{1}+\sum_{k\ge 2}\frac1{(k-2)!}+\color{blue}{\sum_{k\ge 2}\frac1{(k-1)!}}+3\sum_{k\ge 0}\frac1{k!}\\
&=\color{blue}{\sum_{k\ge 0}\frac1{k!}}+4\sum_{k\ge 0}\frac1{k!}\\
&=5\sum_{k\ge 0}\frac1{k!}\\
&=5e\;.
\end{align*}$$
One can also make use of the identity $$x^n=\sum_k{n\brace k}x^{\underline k}\;,$$ where ${n\brace k}$ is a Stirling number of the second kind and $x^{\underline k}\triangleq x(x-1)(x-2)\dots(x-k+1)$ is a falling power. In particular,
$$\begin{align*}
k^3&=\sum_{k=0}^3{3\brace i}k^i\\
&=0\cdot k^{\underline 0}+1\cdot k^{\underline 1}+3\cdot k^{\underline 2}+1\cdot k^{\underline 3}\\
&=k+3k(k-1)+k(k-1)(k-2)\;,
\end{align*}$$
whence 
$$\begin{align*}
\frac{k^2}{(k-1)!}&=\frac{k^3}{k!}=\frac{k+3k(k-1)+k(k-1)(k-2)}{k!}\\
&=\frac1{(k-1)!}+\frac3{(k-2)!}+\frac1{(k-3)!}\;,
\end{align*}$$
and summing over $k$ yields $5e$ as before.
A: 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 $r>0$
$=e+3e+e=5e$
A: I have provided a proof of the general case here.  The proof provided shows
$$
\sum_{k=0}^{\infty} \frac{k^n}{k!} = T_n \cdot e
$$
where $n$ is the $n^{th}$ Bell number.  Yours is the case $n = 3$.  Just for completeness, the formula above is called Dobinski's formula.
A: Let's finish it in one line
$$\sum_{k=1}^\infty \frac{(k-2)(k-1)+3(k-1)+1}{(k-1)!}=5e$$
Chris.
A: Let's start with $e^x=\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!}$ then :
$$x\,e^x=\sum_{k=1}^\infty \frac{x^k}{(k-1)!}$$
$$x(x\,e^x)'=x\,e^x+x^2\,e^x=\sum_{k=1}^\infty \frac{k\,x^k}{(k-1)!}$$
$$x(x(x\,e^x)')'=x(e^x+3x\,e^x+x^2\,e^x)=\sum_{k=1}^\infty \frac{k^2\,x^k}{(k-1)!}$$
Set $x=1$ to get : $$5\,e=\sum_{k=1}^\infty \frac{k^2}{(k-1)!}$$
