Inspired by Surya Prakash

We first note that $\displaystyle\sum_{n=0}^{\infty} x^{n} = \dfrac{1}{1 - x}$ for $|x| \lt 1.$

Differentiating both sides, (the LHS term-by-term), gives us that

$\displaystyle\sum_{n=1}^{\infty} nx^{n - 1} = \dfrac{1}{(1 - x)^{2}}.$

Differentiating once again gives us that

$\displaystyle\sum_{n=2}^{\infty} (n - 1)nx^{n - 2} = \dfrac{2}{(1 - x)^{3}} \Longrightarrow \sum_{n=2}^{\infty} (n - 1)nx^{n - 1} = \dfrac{2x}{(1 - x)^{3}}.$

Now since $|a| \gt 1$ we have that $\dfrac{1}{|a|} \lt 1.$ So plugging in $x = \dfrac{1}{a}$ yields that

$\displaystyle\sum_{n=2}^{\infty} \dfrac{(n - 1)n}{a^{n - 1}} = \dfrac{\dfrac{2}{a}}{\left(1 - \dfrac{1}{a}\right)^{3}} = \boxed{\dfrac{2a^{2}}{(a - 1)^{3}}}.$