Bourbaki's proof of the fundamental theorem of arithmetic 
Theorem $7$ (Decomposition into prime factors). Let $\mathfrak{P}$ denote the set
of prime numbers and $a$ be a strictly positive integer. There exists
one and only one family $(\nu_{p}(a))_{p\in\mathfrak{P}}$ of integers
$>0$ such that the set of $p\in\mathfrak{P}$ with $\nu_{p}(a)\ne 0$ is
finite and $$a=\prod_{p\in\mathfrak{p}} p^{\nu_{p}(a)}.$$
Bourbaki Algebra p.51

Bourbaki says:

...for every family $(\nu_{p}(a))_{p\in\mathfrak{P}}$ satisfying the
conditions of Theorem $7$ $\nu_{p}(a)$ is, for all $p\in\mathfrak{P}$,
equal to the number of factors of a Jordan-Holder series of
$\mathbb{Z}/a\mathbb{Z}$ isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The uniqueness of the family $(\nu_{p}(a))_{p\in\mathfrak{P}}$
therefore follows from the Jordan-Holder theorem.

Can someone please help me understand why this is so? Why does uniqueness follow from Jordan-Holder theorem?
Edit:

A Jordan-Holder series of a group with operators $G$ is a strictly
decreasing composition series $\Sigma$ such that there exists no
strictly decreasing composition series distinct from $\Sigma$ and
finer than $\Sigma$.
Theorem $6$ (Jordan-Holder). Two Jordan-Holder series of a group with
operators are equivalent.

 A: Let $G$ be a group, and let $$G = G_1 \triangleright G_2 \triangleright G_3 \triangleright \cdots \triangleright G_{n} \triangleright G_{n+1} = 1$$ be a composition series of $G$ (Jordan-Hölder series). 
Here $Q_i := G_i/G_{i+1}$ is simple for all $i = 1,\ldots,n$, and $|G| = |Q_1||Q_2| \cdots |Q_{n}|$.
In the case where $G = \mathbb{Z}/a\mathbb{Z}$, you always have $Q_i \cong \mathbb{Z}/p_i\mathbb{Z}$ for some prime $p_i$, and so $a = p_1p_2 \cdots p_n$.

Conversely when $a = p_1p_2 \cdots p_{n}$ for primes $p_i$, you can construct a composition series of $G = \mathbb{Z}/a\mathbb{Z}$ with $Q_i \cong \mathbb{Z}/p_i\mathbb{Z}$.
Hence if $a = p_1 p_2 \cdots p_n = p_1' p_2' \cdots p_m'$ for some primes $p_i, p_j'$, by Jordan-Hölder $m = n$ and $\{p_1, \ldots, p_n\} = \{p_1', \ldots, p_m'\}$.
A: This is an immediate consequence of the Jordan-Holder Theorem, e.g. see below from Rotman's Advanced Modern Algebra, I, (3ed), p. 198.

Remark $ $ The basic "Schreier refinement theorem" holds more generally for modular lattices. Such lattice-theoretic or refinement-based view of (unique) factorization prove useful in many applications, e.g. it proved quite fruitful in Paul Cohn's work on generalizations of unique factorization to noncommutative rings. 
To learn more about the various related viewpoints follow the links regarding that and Schreier rings and Riesz interpolation in the Remark in this answer (about the high-school AC-method for reducing polynomial factorization to the monic case).
