proof of formula for increasing annuity

The elegance of the solution comes from comparing the series with a shifted version of itself; subtraction removes most of the terms.

How would someone arrive at this solution under their own efforts:

by understanding typical algebraic technique of aligning two similar but messy expressions

the proof

an increasing immediate annuity pays: \(1 + 2 + 3 + \cdots + n\) at times \(1 + 2 + 3 + \cdots + n\) respectively:
\(\begin{align} Ia_{\overline{n}|} = \sum_{k=1}^{n} kv^k \tag{1} \end{align}\)

\(= v + 2v^2 + 3v^3 \cdots + (n-1)v^{n-1} + nv^{n}\)

and

\(vIa_{\overline{n}|} = v^2 + 2v^3 + 3v^4 \cdots + (n-1)v^{n} + nv^{n+1}\)

so that: \((1-v)Ia_{\overline{n}|} = v^2 + v^3 + \cdots + v^{n-1} + v^n - nv^{n+1}\)

an immediate annuity: \(a_{\overline{n}|} = \sum_{k=1}^{n} v^k\)