Why is this equation $\sum_{k=0}^{\infty}k^2\frac{\lambda^k}{k!e^\lambda}=\lambda +\lambda^2$ true? Why is $$\sum_{k=0}^{\infty}k^2\frac{\lambda^k}{k!e^\lambda}=\lambda +\lambda^2$$
For the context: I am trying to calculate $E(X^2)$, where X is a poisson distributed random variable.
All my calculations lead to a dead end. Is there a trick to process the $k^2$? The only thing I see worth doing is pulling out $1/e^\lambda$.
Edit: Considering @Srivatsan's hint I got:
$$\sum_{k=0}^{\infty}k^2\frac{\lambda^k}{k!e^\lambda}=e^{-\lambda}\sum_{k=0}^{\infty}(k(k-1)+k)\frac{\lambda^k}{k!}=e^{-\lambda}\left( \sum_{k=0}^{\infty}k(k-1)\frac{\lambda^k}{k!}+\sum_{k=0}^{\infty}k\frac{\lambda^k}{k!}\right)$$
$$=e^{-\lambda}\sum_{k=0}^{\infty}k(k-1)\frac{\lambda^k}{k!}+e^{-\lambda}\sum_{k=0}^{\infty}k\frac{\lambda^k}{k!}=e^{-\lambda}\lambda^2\sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{(k-2)!}+e^{-\lambda}\lambda\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!}$$
$$=e^{-\lambda}\lambda^2\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}+e^{-\lambda}\lambda\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}$$
$$=\lambda^2+\lambda$$
And here we are! Thank you very much, @Srivatsan!
 A: HINT: Write $k^2 = k(k-1) + k$, and split the sum into two. [Update: The OP has added the complete solution to the post.] 

If you are interested in a general $d^{th}$ moment of this distribution, then we are faced with an expression of the form
$$
\mathrm e^{-\lambda} \sum_{k=0}^{\infty} k^d \frac{\lambda^k}{k!}. 
$$
To proceed, we first write $x^d$ as a linear combination of the $d$ falling factorial polynomials  (also called Pochhammer symbols): 
$$
\begin{align*}
(x)_1 &= x  
\\
(x)_2 &= x(x-1) 
\\
(x)_3 &= x(x-1)(x-2)
\\
&\vdots
\\
(x)_d &= x(x-1)(x-2) \cdots (x-d+1) 
\end{align*}
$$
[I am following the notation used in the wikipedia article.] After this step, the rest of the manipulations is very similar to the $d=2$ case that the OP showed in the question. 
The falling factorial polynomials appear in multiple contexts. First, the Pochhammer symbol $(x)_i$ is obviously related to the binomial coefficient $\binom{x}{i}$. Further,  as seen above, these polynomials are very useful in manipulating summations. See also this answer of robjohn. 
A: The moments of the Poisson distribution, as functions of the first moment, are given by the Touchard polynomials, also called "exponential polynomials".  Their coefficients are related to the enumeration of set partitions.  See Dobinski's formula.
