Proof: By Induction

By hypothesis, $a,b\in R$ are elements of a unit ring $R,$ and $n\in\mathbb N$ is a natural number.
We provide a proof by induction on $n.$
Base case: $n=0$ $$(a-b)\sum_{k=0}^0 a^k\cdot b^{-k}=(a-b)\cdot 1=a^1-b^1.$$
Induction step $n\to n+1$
- Assume, $$(a-b)\sum_{k=0}^n a^k\cdot b^{n-k}=a^{n+1}-b^{n+1}$$ holds for some $n\ge 0.$
- Then $$\begin{align}(a-b)\sum_{k=0}^{n+1} a^k\cdot b^{n+1-k}&=(a-b)\sum_{k=0}^{n} a^k\cdot b^{n+1-k}+(a-b)\cdot ( a^{n+1}b^{0})\nonumber\\ &=b(a-b)\sum_{k=0}^{n} a^k\cdot b^{n-k}+a^{n+2}-a^{n+1}b\nonumber\\ &=b(a^{n+1}- b^{n+1})+a^{n+2}-a^{n+1}b\nonumber\\ &=ba^{n+1}- b^{n+2}+a^{n+2}-a^{n+1}b\nonumber\\ &=a^{n+2}-b^{n+2}\nonumber\\ \end{align}$$
Thus, the geometric sum holds for all $n\in\mathbb N.$
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Modler, F.; Kreh, M.: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition