TITLE: Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit
QUESTION [11 upvotes]: Prove $\alpha \in R[[x]]$ is a unit iff $a_0 \in R$ is a unit.

$"\Rightarrow"$ suppose $\alpha \in R[[x]]$ is a unit, where $\alpha = \sum_{n=0}^{\infty} a_nx^n$
then there exists an inverse $\beta = \sum_{n=0}^{\infty} b_nx^n$ 
stuck at this spot,
$(\sum_{n=0}^{\infty} a_nx^n)(\sum_{n=0}^{\infty} b_nx^n) = \sum_{n=0}^{\infty}(\sum_{k=0}^{n}(a_kb_{n-k}))x^n$
A little mixed up on how this becomes $1 = \sum_{n=0}^{\infty}e_nx^n$ where $e_0 =1, e_{n>0} = 0$ and why that would make $a_0b_0 = 1$ which would make $a_0 \in R$ a unit. I haven't tried the other direction yet.
EDIT
Maybe I get it, because we assumed this sum is a unit then its inverse exists and so the $x^0=1$ and all $a_ib_i=1$ thus $a_0b_0=1$ which gives $a_0$ is a unit in $R$

REPLY [4 votes]: $\alpha \beta =1$ means $(a_0 + a_1 x+\cdots)(b_0 + b_1 x+ \cdots)=1+0x+\cdots$ so $a_0b_0+ (a_1b_0+a_0 b_1)x+\cdots=1+0x+\cdots$ therefore $a_0b_0=1$