In a ring, $1-ab$ is invertible $\implies$ $1-ba$ is invertible (from scratch)

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I know the usual trick: let $c=(1-ab)^{-1}$ and notice that:

$$(1-ba)(bca) = bca - babca = b(c-abc)a=bc(1-ab)a=ba$$

Etc.

I was wondering how to discover this fact from scratch, without invoking this magical observation. Any help in that?