I want to show that the norm $N_{K/\mathbb Q}(\mathfrak{a})$ of $\mathfrak{a}$ a nonzero integral ideal of a number field $K$ is finite, and so $N_{K/\mathbb Q}(\mathfrak{ab})=N_{K/\mathbb Q}(\mathfrak{a})N_{K/\mathbb Q}(\mathfrak{b})$ for both $\mathfrak{a,b}$ nonzero integral ideals.
I know that the norm of a nonzero integral ideal of the ring of integers $\mathcal{O}_K$ is defined as the index of $\mathfrak{a}$ in $\mathcal{O}_K$, ie. $N_{K/\mathbb Q}(\mathfrak{a})=[\mathcal{O}_K:\mathfrak{a}]=|(\mathcal{O}_K/\mathfrak{a})|$.
I also know that, for $\alpha \in \mathcal{O}_K^*$, $|N_{K/\mathbb Q}(\alpha)|=[\mathcal{O}_K:\mathcal{O}_K\alpha]$ but I don't think this is very useful considering nonzero integral ideals specifically.
I'm not sure how to tackle this. I assume, once considering the definition of the norm, the problem is reduced to proving the properties for the fields $(\mathcal{O}_K/\mathfrak{a})$ and $(\mathcal{O}_K/\mathfrak{b})$, but how do I prove any results using them? I know that the ring of integers $\mathcal{O}_K$ is finitely generated, but I'm not sure if that is useful either.
For the finiteness you should use that $\mathcal O_k \cong \mathbb Z^n \cong \mathfrak a$ as abelian groups, this implies that the quotient is finitely generated torsion group, hence finite.
For the multiplicativity you should notice that the chinese remainder theorem ensures that it suffices to show $N(\mathfrak p^n) = a^n$, when $N(\mathfrak p)=a$.
To that account consider the exact sequence
$$0 \to \mathfrak p^{n-1}/\mathfrak p^n \to \mathcal O_k/\mathfrak p^n \to \mathcal O_K/\mathfrak p^{n-1} \to 0$$
and use induction.