Solution for internal use NOT for publication
We know that

\sum_{n = 0}^{\infty} x^{n} = \frac{1}{(1 - x)}

\begin{matrix} \sum_{n = 0}^{\infty} {nx}^{n} & = x \frac{d}{dx} \sum_{n = 0}^{\infty} x^{n} \\ = x \frac{d}{dx} (\frac{1}{1 - x}) \\ = \frac{x}{(1 - x)^{2}} \end{matrix} .

Going one step further

\begin{matrix} \sum_{n = 0}^{\infty} n^{2} x^{n} & = x \frac{d}{dx} \frac{x}{(1 - x)^{2}} \\ = x (\frac{(1 - x)^{2} + 2 x (1 - x)}{(1 - x)^{4}}) \\ = x (\frac{(1 - x) + 2 x}{(1 - x)^{3}}) \\ = \frac{x (1 + x)}{(1 - x)^{3}} \end{matrix} .

More generally

$\sum_{n = 0}^{\infty} n^{k} x^{n} = (x \frac{d}{dx}) \dots (x \frac{d}{dx}) (\frac{1}{1 - x})$

where the operator.

$(x \frac{d}{dx})$

is applied $k$ times.