If $R$ is an integral domain, then $R[[x]]$ is an integral domain While solving another problem (specifically Exercise 7.2 in Atiyah & Macdonald's Introduction to  Commutative Algebra), I got stuck in the following step:

If $R$ is an integral domain, how I can prove that $R[[x]]$ is an
  integral domain?

Here $R[[x]]$ is the set of all formal series in $x$ with coefficients in $R$. So typical element of $R[[x]]$ would have the form $a_0+a_1x+a_2x^2+\cdots$ where $a_i\in R$.
So I need to prove that if
$$
(a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)=0
$$
then $a_i=0$ and $b_i=0$ for all $i\ge 0$. Now, I am not particularly fond of opening up those brackets :( Is there any slick way of proving this?

Since $(a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)=0$, the term of degree $0$ must be zero. Therefore the constant coefficient $a_0b_0=0$, assume $b_0 \ne 0$. Then $a_0=0$ and the $x$ coefficient is $a_1b_0$ so $a_1=0$. Now try to find some strong induction to show for any $n$, $a_n=0$.
