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?

Notice we only have to prove that if the $b_i$ are not all $0$, then the $a_i$ are all $0$.
Just use induction.  First, let $m$ be minimal with $b_m\ne 0$, so that $a_0=0$ by looking at the coefficient of $x^m$ in the product.  Next, suppose that $a_k=0$ for all $k=0,\ldots,n$, and notice that the coefficient of $x^{m+n+1}$ in the product is $a_{n+1}b_m$.  Since $b_m\ne 0$, it follows that $a_{n+1}=0$.
