I am learning ring theory in the Dummit & Foote's Abstract Algebra, and I am doing all the exercises to get as much experience as possible... but some of them just get me stuck for hours! Like this one :

*Assume $R$ is a commutative ring with identity. Prove that $p(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in R[x]$ is a unit if and only if $a_0$ is a unit and $a_1, \dots, a_n$ are nilpotent in $R$.*

Now I already have the $\Longleftarrow$ part, since I've proven in a previous exercise that the sum of nilpotent elements is nilpotent and the sum of a unit with a nilpotent element is a unit, hence $a_0 + (a_1 x + \dots + a_n x^n)$ has a sum of nilpotents between the parenthesis, and thus is written as the sum of a unit and a nilpotent element, thus is a unit. I can't deal with the converse though.

I've tried noticing that if $p(x)$ is a unit and $q(x)$ is its inverse then $p(x) q(x) = 1$ implies that $p(x)^m q(x)^m = 1$ (because $R$ is commutative), but that only gave me that if $p$ has degree $n$ and $q$ has degree $r$ with last coefficient $b_r$ then $a_n b_r$ is $0$, and got stuck there. I also tried to take a look at what $p(x) q(x)$ looks like : If $p(x) = \sum_{k=0}^n a_k x^k$ and $q(x) = \sum_{k=0}^n b_k x^k$ (just add zeros to get those two sums of same size), then $$ a_0 b_0 =1 ,\quad a_1 b_0 + a_0 b_1 = 0, \quad a_2 b_0 + a_1 b_1 + a_0 b_2 = 0, \quad \cdots $$ and I thought I could get something out of those equations but all I have now is that $a_1$ is nilpotent if and only if $b_1$ is, which doesn't help me much.

Any hints? I don't need a full solution if there's a way to just point out a nice trick.