This is Lemma 11.23 in Atiyah:
For an ideal $\mathfrak a \subseteq A$, define $G_{\mathfrak a} (A) = \bigoplus _{n=0} ^\infty \mathfrak a^n / \mathfrak a^{n+1}$.
The statement of the Lemma:
Let $A$ be a ring, $\mathfrak a$ an ideal of $A$ such that $\bigcap_n \mathfrak a^n = 0$. Suppose that $G_{\mathfrak a} (A)$ is an integral domain. Then $A$ is an integral domain.
Proof: Let $x,y$ be nonzero elements of $A$. Then since $\bigcap \mathfrak a^n = 0$, there exist nonnegative integers $r,s$ such that $x \in \mathfrak a^r - \mathfrak a^{r+1}$ and $y \in \mathfrak a^s - \mathfrak a^{s+1}$.
My question is why should the powers of $\mathfrak a$ cover everything in $A$ to begin with? Certainly if this were true, and the intersection of all the powers is zero, then the above follows. So it seems that this depends on something along the lines of
$$\bigcup \mathfrak a^n = A.$$
Does it have to do with the $\mathfrak a$-adic topology where these powers of $\mathfrak a$ are the neighborhoods of $0$?
The numbers $r$ and $s$ are allowed to be $0$, in which case by convention we define $\mathfrak a^0=A$. This makes sense, because $\mathfrak a^0$ should be the ideal generated by all $0$-fold products of elements of $\mathfrak a$, and the only such product is the empty product, $1$. The ideal generated by $1$ is all of $A$.
Note that if you did not allow $r$ and $s$ to be $0$, this would almost always be false. Indeed, $\mathfrak a^n\subseteq a$ for all $n>0$ since $\mathfrak a$ is an ideal, so $\bigcup_{n>0}\mathfrak a^n$ is just $\mathfrak a$.