If we look at page $66$ on Marcus' Number Fields, we find that we are trying to prove, among other things, that $\left \|IJ\right \|=\left \|I\right \|\left \|J\right \|$ for nonzero ideals $I$ and $J$ of a number ring $R$, where $\left \|I\right \|:=\left |R/I\right |$.
Since $R$ is a number ring, it is a Dedekind Domain, therefore we have prime decomposition of ideals. We start by supposing that $I$ and $J$ are relatively prime, in which case the assertion is obvious by the chinese remainder theorem and the fact that $IJ=I\cap J$.
In order to finish, we want $\left \|P^m\right \|=\left \|P\right \|^m$ for $m\in \mathbb{N}$ and $P$ a prime ideal in $R$. Since we have a chain $P^{i+1}\subset P^i$ for $i\in \left [0,m-1\right ]\cap \mathbb{Z}$, Marcus says that it suffices to show that $\left \|P\right \|=\left |P^i/P^{i+1}\right |$, where the $P^i$ are considered as abelian subgroups. At this step I got stuck, since I cannot see why it is sufficient.
I'll show that $\;\bigl\lVert \mathfrak p\bigr\rVert=\bigl\lvert\mathfrak p/\mathfrak p^2\bigr\rvert$ is enough to prove that $$\;\bigl\lVert \mathfrak p^2\bigr\rVert=\bigl\lVert \mathfrak p\bigr\rVert^2.$$
Indeed the short exact sequence: $$0\longrightarrow \mathfrak p/\mathfrak p^2 \longrightarrow R/\mathfrak p^2 \longrightarrow R/\mathfrak p \longrightarrow 0$$ shows that $$\bigl\lVert \mathfrak p^2\bigr\rVert=\bigl\lvert\mathfrak p/\mathfrak p^2\bigr\rvert\cdot\bigl\lvert R/\mathfrak p\bigr\rvert=\bigl\lvert\mathfrak p/\mathfrak p^2\bigr\rvert\cdot\bigl\lVert \mathfrak p\bigr\rVert = \bigl\lVert \mathfrak p\bigr\rVert^2.$$ The general case follows by an easy induction, considering the short exact sequence: $$0\longrightarrow \mathfrak p^i/\mathfrak p^{i+1} \longrightarrow R/\mathfrak p^{i+1} \longrightarrow R/\mathfrak p^i \longrightarrow 0$$