= Solution
Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the <cyclic group> of order $n$: it has a chain of subgroups of successive orders
$$
n,\quad n/p_1,\quad n/(p_1p_2),\quad\ldots,\quad1.
$$
Read in increasing order, this is a <composition series> with factors isomorphic to the cyclic groups of orders $p_1,\ldots,p_r$. Any second prime factorization of the same integer produces another <composition series> of the same group $C_n$.
The <Jordan–Hölder theorem> equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. Hence
$$
\boxed{\text{The prime factorization is unique up to permutation}.}
$$
For $n=1$ both factor lists are empty, corresponding to the trivial group.
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