If $u_1u_2\cdots u_n=1$ in a commutative ring, then all of $u_i$ are units If $u_1u_2\cdots u_n=1$ in a commutative ring, then all of $u_i$ are units.
Does the proof follow some logic like the following:
$u_1(u_2\cdots u_n)=1\implies u_1,u_2\cdots u_n$ are both units, so $u_1$ is a unit,
$u_2(u_1u_3u_4\cdots u_n\implies u_2,u_1u_3\cdots u_n)$ are both units, so $u_2$ is a unit,
etc
Or is there some other way to show this?

Here is a proof by induction.
The cases $n=1$ and $n=2$ are clear.
If $n>2$, let $v=u_1u_2$. Then $vu_3\cdots u_n=1$ and by induction $v, u_3, \ldots, u_n$ are units. Use the case $n=2$ on $v$ to conclude that $u_1, u_2$ are units.
