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Use $[g]$ for the basis element of the group algebra $A={\mathbb{F}}_p[{\mathbb{Z}}_p]$, so $[g][h]=[g+h]$ and $[0]=1$. The kernel of the augmentation map is the augmentation ideal $I$. This algebra must not initially be identified with a Laurent polynomial ring: the additive group ${\mathbb{Z}}_p$ is not abstractly generated by the integer $1$.

For $r\ge 1$, map $A$ onto ${\mathbb{F}}_p[{\mathbb{Z}}_p/p^r{\mathbb{Z}}_p]$ and call its kernel $J_r$. It is generated by $[g]-1$ for $g\in p^r{\mathbb{Z}}_p$: differences of basis elements in one coset have the form $[h]([g]-1)$. Since each such $g=p^{r}h$,by the characteristic-$p$ binomial identity. Thus $J_r\subseteq I^{p^r}$.

The quotient group is a cyclic group of order $p^r$, generated by the image of $1$. Its group algebra isThe image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. HenceThis is cyclic finite quotients of the p-adic augmentation filtration.

For any $n\ge 0$, choose $p^r>n+1$. The finite quotient then identifies $I^n/I^{n+1}$ with $(T^n)/(T^{n+1})$, a one-dimensional space generated by the image of $([1]-1{)}^n$. The degree-one symbol $X$ of $[1]-1$ consequently gives a surjective graded map ${\mathbb{F}}_p[X]\to {\mathrm{gr}}_{I}A$, and no power of $X$ vanishes in its degree. ThusThe same finite quotients show that the I-adic completion is ${\mathbb{F}}_p[[T]]$, and its associated graded ring is identical. The proof therefore covers the completed group algebra interpretation too, while proving the literal ordinary-algebra statement without assuming completion. Applying part ii to a ring would require its actual completeness, not just this polynomial associated graded ring.