= Solution
Using the assumed <square-free polynomial> property, factor
$$
X^p-1=f_1\cdots f_r
$$
into distinct <monic polynomials> that are <irreducible polynomials>, with $n_j=\deg f_j$. Part (b)(iii) identifies $T=\mathbb F_2[X]/(X^p-1)$ with $\prod_jK_j$, where $|K_j|=2^{n_j}$.
By the <first isomorphism theorem for rings>, the image $S$ from part (c)(i) is $T/J$, where $J$ is the <kernel of a ring homomorphism>. By part (a)(ii), $J=\prod_jJ_j$ for <ideals> $J_j\subseteq K_j$. A <field> has only the zero <ideal> and the whole <field> as <ideals>, so each $K_j/J_j$ is either $K_j$ or the <zero ring>. Consequently, for a subset $D\subseteq\{1,\ldots,r\}$,
$$
S\cong\prod_{j\in D}K_j.
$$
This is the <quotient of a finite product of fields> principle. The subset is nonempty because $|U(S)|=p>2$, whereas the <zero ring> has only one <unit>. By part (a)(i),
$$
\boxed{p=|U(S)|=\prod_{j\in D}(2^{n_j}-1)}.
$$
Each factor is a positive <integer>. Since $p$ is a <prime number>, exactly one factor is $p$ and all remaining factors are one. Thus
$$
\boxed{p=2^n-1\text{ for some }n\geq2},
$$
so $p$ is a <Mersenne prime>. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a <prime number>, although only the displayed form is needed.
The assumed <square-free polynomial> property can also be checked directly: in <characteristic of a ring> two, the <formal derivative> of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two <polynomials> have <greatest common divisor> one, since $X$ does not divide $X^p-1$, so no repeated <irreducible polynomial> factor occurs.
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