3 \cdots

If $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$.

Proof: $$(1-ba)^{-1} = 1+ba +baba+... baba+\cdots = 1+b(1+ab+abab+..)a 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$
The meaningless infinite series give the right answer (which is hard to guess).

2 just latexified some things for completeness

If 1-ab $1-ab$ is invertible for a, b $a$, $b$ in a (noncommutative) ring then so is 1-ba. $1-ba$.

Proof: $(1-ba)^{-1}$(1-ba)^{-1} = 1+ba +baba+... = 1+b(1+ab+abab+..)a = 1+b(1-ab)^{-1}a$,1+b(1-ab)^{-1}a,$$The meaningless infinite series give the right answer (which is hard to guess). 1 [made Community Wiki] If 1-ab is invertible for a, b in a (noncommutative) ring then so is 1-ba. Proof:$(1-ba)^{-1} = 1+ba +baba+... = 1+b(1+ab+abab+..)a = 1+b(1-ab)^{-1}a\$,
The meaningless infinite series give the right answer (which is hard to guess).