I want to prove the statement that completion of a ring w.r.t. maximal ideal is local. Let $\mathfrak{m}$ be a maximal ideal of a ring $A$, and $\hat{A}$ be an $\mathfrak{m}$-adic completion of $A$.
I know that my statement can be verified if an element $$X:=1+\Sigma_{i\geq1} x_i $$in $\hat{A}$is unit in $\hat{A}$ ($x_i \in \mathfrak{m}^i$). Using the geometric series, we should verify that $$1/X = 1/\{{1-(1-X)}\} = \Sigma(1-X)^i$$ is an element in $\hat{A}$. But I don't know why this element is in $\hat{A}$, because even though a general element in $\hat{A}$ can be represented by infinite series with each term lies in $\mathfrak{m}^i$, but $$1/X = \Sigma(1-X)^i = 1-(x_1+x_2+\cdots)+(x_1+x_2+\cdots)^2-\cdots $$ contains infinite operation "in $\mathfrak{m}^i$", which is absurd. Can anyone tell me why we can say that $1/X$ is in $\hat{A}$?