Elements congruent to $1$ modulo a defining ideal are units

For each $n\ge 1$, the finite geometric sum $v_n:=1+a+a^2+\cdots +a^{n-1}$ satisfies $(1-a)v_n=1-a^n\mathrm{.}$ Since $a\in I$, one has $a^n\in I^n$, so the image of $v_n$ in $R\mathrm{/}{I}^n$ is an inverse to the image of $u=1-a$.

The residue classes $(v_n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>I^n{)}_n$ are compatible: the image of $v_{n+1}$ in $R\mathrm{/}{I}^n$ equals the image of $v_n$ because $v_{n+1}-v_n=a^n\in I^n$. Therefore they define an element

By [L1], there is a unique element $w\in R$ corresponding to $v$. Since each component of $v$ is an inverse to the image of $u$, the products $uw$ and $wu$ map to $1$ in every quotient $R\mathrm{/}{I}^n$. Completeness includes separatedness, so the kernel of is $0$; hence $uw=wu=1.$ Thus $u$ is a unit.

Let $x\in I$ and $r\in R$. Then $1-rx\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}I)$, so step 3.1 shows $1-rx$ is a unit. The elementary ideal characterization of the Jacobson radical now gives $x\in J(R)$: if a maximal ideal omitted $x$, its image would generate the residue field, contradicting invertibility of every $1-rx$. Therefore $I\subseteq J(R)$.