If $r,s $ are elements of a ring, then $1-rs$ invertible implies $1-sr$ is invertible (and it is a trick: you can make an educated guess for the formula for the inverse of $1-sr$ from that for $1-rs$). This can be used to find quick proofs of: (a) in a Banach algebra, ${\rm spec\ } rs \cup \{0\} = {\rm spec}\ sr \cup \{0\}$ (which in turn yields the nonsolvability of $xy-yx = 1$---all one needs is boundedness and nonemptiness of the spectrum); (b) the Jacobson radical (defined as the intersection of all maximal right ideals) is a two-sided ideal; and probably some other things I can't think of right now ...