All bases for a finitely-generated abelian group have the same cardinality.

All bases for a finitely-generated abelian group have the same cardinality.

I want to understand more about this proof from Lang's Algebra:
Let $B$ be a subgroup of a free abelian group $A$ with basis
$(x_i)_{i=1...n}$. It has already been shown that $B$ has a basis of
cardinality $\leq n$.
... We also observe that our proof shows that there exists at least one
basis of $B$ whose cardinality is $\leq n$. We shall therefore be finished
when we prove the last statement, that any two bases of $B$ have the same
cardinality. Let $S$ be one basis, with a finite number of elements $m$.
Let $T$ be another basis, and suppose that $T$ has at least $r$ elements.
It will suffice to prove that $r \leq m$ (one can then use symmetry).
Let $p$ be a prime number. Then $B/pB$ is a direct sum of cyclic groups of
order $p$, with $m$ terms in the sum. Hence its order is $p^m$. Using the
basis $T$ instead of $S$, we conclude that $B/pB$ contains an $r$-fold
product of cyclic groups of order $p$, whence $p^r \leq p^m$ and $r \leq
m$, as was to be shown. (Note that we did not assume a priori that T was
finite.)
I've bolded the parts I'm having trouble with. How do I show the first
part and what's an $r$-fold product?
Alternative proofs to the problem welcome.
I know that $pB = \{ \sum_{i} p k_i x_i\ | \sum_{i} k_i x_i \in B\}$ and
that it forms a normal subgroup, $A$ being abelian.