= Solution
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 $r\ge1$, 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 $g=p^rh$,
$$
[g]-1=([h]-1)^{p^r}\in I^{p^r}
$$
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> is
$$
\mathbb F_p[u]/(u^{p^r}-1)=\mathbb F_p[T]/(T^{p^r}),\qquad T=u-1.
$$
The image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. Hence
$$
\boxed{J_r=I^{p^r},\qquad A/I^{p^r}\cong\mathbb F_p[T]/(T^{p^r})}.
$$
This is <cyclic finite quotients of the p-adic augmentation filtration>.
For any $n\ge0$, 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 $\mathbb F_p[X]\to\operatorname{gr}_IA$, and no power of $X$ vanishes in its degree. Thus
$$
\boxed{\operatorname{gr}_I\mathbb F_p[\mathbb Z_p]\cong\mathbb F_p[X]}.
$$
The 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>.
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