E(X) in Poisson Dist,
\begin{align} \mathrm{E}(X) &= \sum_{k=0}^{\infty} \frac{k \lambda^k e^{-\lambda}}{k!}= \\ &= e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^{k+1}}{k!}=\\ &= \lambda \end{align}
then,
$\text{E}(X(X-1)) = \sum_{k=0}^{\infty} \frac{k(k-1) \lambda^{k(k-1)} e^{-\lambda}}{k(k-1)!}$
how can I simplify this ?
Probability generating functions are probably the easiest route. The PGF for $X$ is $$ G_X(z)=\sum_{k=0}^{\infty}\frac{e^{-\lambda}\lambda^k}{k!}z^k=e^{\lambda(z-1)}. $$ Then, use the fact that $\mathbb{E}[X(X-1)]=G''(1)$.
Also, note that your expression for $\mathbb{E}[X(X-1)]$ is incorrect: it should be $$ \mathbb{E}[X(X-1)]=\sum_{k=0}^{\infty}k(k-1)P(X=k)=\sum_{k=0}^{\infty}k(k-1)\frac{e^{-\lambda}\lambda^k}{k!}. $$ If you are uncomfortable with PGFs, you could also work directly here by noting that the $k=0$ and $k=1$ terms are 0, and for $k\geq2$ we have $$ \frac{k(k-1)}{k!}=\frac{1}{(k-2)!}, $$ so that $$ \mathbb{E}[X(X-1)]=\lambda^2e^{-\lambda}\sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{(k-2)!}=\lambda^2e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}=\lambda^2. $$
Obviously, this works just fine; however, you really should try to get comfortable with PGFs if you aren't already - they are extremely helpful tools!
We want $\sum_{k=0}^\infty k(k-1)e^{-\lambda}\frac{\lambda^k}{k!}$.
Note that the terms for $k=0$ and $k=1$ are $0$. So we can sum from $2$ on. Then use the fact that $\frac{k(k-1)}{k!}=\frac{1}{(k-2)!}$. So the expectation is $$\lambda^2 \left(\sum_{k=2}^\infty e^{-\lambda}\frac{\lambda^{k-2}}{(k-2)!}\right).$$
Let $j=k-2$. Our expectation is $$\lambda^2\left(\sum_{j=0}^\infty e^{-\lambda}\frac{\lambda^j}{j!}\right).$$ The sum in parentheses above is $1$, so our expectation is $\lambda^2$.
Remark: Note that this generalizes immediately to $X(X-1)(X-2)$, $X(X-1)(X-2)(X-3)$, and so on. As a general heuristic, these kinds of expressions often behave more nicely than powers.
