My question is that $f= a_0 + a_1X + ... + a_nX^n\in A[x]$ is unit with inverse $g= b_0 + b_1X + ... + b_mX^m$
My Question in the first induction step $a_nb_m=0$ why it implies $a_n$ nilpotent?
if $a_n$ and $b_m$ are zero divisors then $a_nb_m$ is zero.
For example consider: $\mathbb Z_6$, we can see that $\bar 2 * \bar 3=\bar 0$ but $\bar 2^n\neq 0$ for any $n$. (Since any power of $2$ cannot $=0 \quad mod 6=2.3$, $2^n$ never divisible by $3$)
In your link (which you deleted from your post), you have to read not only the answer (which focuses on a detail, as requested) but the question (which describes the whole proof). The first induction step is indeed $a_nb_m=0$, but the last one is $a_n^mb_0=0$, and it is this last step which (since $b_0$ is invertible) proves that $a_n$ is nilpotent.
I hear you grumble, like I did: "ok then, $a_n$ is nilpotent, but what about $a_{n-1},\dots,a_1$?" The key is in N. S. answer's in the duplicate of your link.
