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You can just leave $q$ as an indeterminate; note that the expression is $$ q^0+q^1+\dots+q^n $$ and this means that, for $n=0$, it is just $q^0=1$. So the base step is true. Jun 13, 2020 · Is there a formal proof for $(-1) \\times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?
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