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?

No induction required. 
If there exists  $v$ such that $(u_1\dotsm u_n)v=1$, then $u_2\dotsm u_nv $ is the inverse of $u_1$. Similarly for all other $u_i$s.
