I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.

**Theorem.** Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all *nilpotent* in $R$.

**Proof.** Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.

I have not been able to get the converse part and would like to see the proof for the converse part.