Exponential identity How can one show the following identity 
$ne^n = \sum_{k=0}^{\infty}n^k(n - k)^2/k!$ ?
 A: Expanding on Didier's answer.
It is convenient to replace $n$ by $\lambda$. Then, you want to show that
$$
\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}}  = \lambda e^\lambda . 
$$
Starting as Didier suggested, write the left-hand side as
$$
\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}}  = \lambda ^2 e^\lambda  \sum\limits_{k = 0}^\infty  {e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  - 2\lambda e^\lambda  \sum\limits_{k = 0}^\infty  {ke^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  + e^\lambda  \sum\limits_{k = 0}^\infty  {k^2 e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}.
$$
Now, if $X$ is a Poisson random variable with mean $\lambda$, then
$$
{\rm E}(X) = \sum\limits_{k = 0}^\infty  {ke^{ - \lambda } \frac{{\lambda ^k }}{{k!}}}  
$$
and
$$
{\rm E}(X^2) = \sum\limits_{k = 0}^\infty  {k^2 e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}} .
$$
From the facts that ${\rm E}(X) = \lambda$ and ${\rm E}(X^2) = {\rm Var}(X) + {\rm E}^2 (X) = \lambda + \lambda^2$ (and noting that $\sum\nolimits_{k = 0}^\infty  {e^{ - \lambda } \frac{{\lambda ^k }}{{k!}}} = 1$), it follows that
$$
\sum\limits_{k = 0}^\infty  {\frac{{\lambda ^k (\lambda  - k)^2 }}{{k!}}} = e^\lambda  (\lambda ^2  - 2\lambda ^2  + \lambda  + \lambda ^2 ) = \lambda e^\lambda .
$$
A: By developing the square as $(n-k)^2=n^2-(2n-1)k+k(k-1)$ and evaluating each of the three resulting sums over $k$.
