Question: 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?

Answer: 
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.
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.