On the norm formula $N(IJ) = N(I)N(J)$ in an order of an algebraic number field

Let $K$ be an algebraic number field of degree $n$.
Let $\mathcal{O}_K$ be the ring of algebraic integers of $K$.
Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $n$.
Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$.
Let $I$ be a non-zero ideal of $R$.
If $I + \mathfrak{f} = R$, we call $I$ regular.
For the properties of regular ideals, see this.
We denote by $N(I)$ the number of elements of the finite ring $R/I$.

If $I$ and $J$ are regular ideals of $R$, we have $N(IJ) = N(I)N(J)$ as proved in my answer to this question.
I wonder if the formula still holds when one of them is not regular.
And I came up with the following proposition.

Proposition
Let $R$ be an order of an an algebraic number field.
Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$.
Then $N(IJ) = N(I)N(J)$.

Outline of my proof
I generalized the result of this and proved $R/J$ is $R$-isomorophic to $I/IJ$.
See my answer below for the detail.

My question
How do you prove the proposition?
I would like to know other proofs based on different ideas from mine.
I welcome you to provide as many different proofs as possible.
I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.

Solutions Collecting From Web of "On the norm formula $N(IJ) = N(I)N(J)$ in an order of an algebraic number field"

Lemma
Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$.
Then there exists non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = A$.

Proof:
By this question, $I$ is uniquely decomposed as a product of regular prime ideals.
Let $I = P_1^{e_1}…P_n^{e_n}$ be the prime decomposition, where $P_1,\cdots,P_n$ are distinct regular prime ideals and $e_i \ge 1$ for all $i$.
Let $Q_1, …, Q_m$ be all the distinct prime ideals which contain $J$, but not contain $I$.
For each $i$ choose $\alpha_i\in\ P_i^{e_i}\setminus P_i^{e_i+1}$.
For each $j$ choose $\beta_j\in\ A\setminus Q_j$.
Note that $P_1,\cdots,P_n$ and $Q_1, …, Q_m$ are maximal ideals.
Hence, by the Chinese Remainder Theorem, there exists $\alpha\in A$ such that
$\alpha\equiv\alpha_i\pmod{P_i^{e_i+1}}$ for all $i$ and
$\alpha\equiv\beta_j\pmod{Q_j}$ for all $j$.
Since $\alpha \in P_i^{e_i}$ for all $i$, $\alpha \in I$.
Since $I$ is invertible by this question, there exists an ideal $M$ such that $\alpha R = IM$.
Clearly $M + J = A$

Proof of the proposition
It suffices to prove that $R/J$ is isomorphic to $I/IJ$.
By the lemma, there exist non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = R$.
Then $IM + IJ = I$.
Let $\psi\colon R \rightarrow I/IJ$ be the $R$-homomorphism sending $x$ to $\alpha x$ (mod $IJ$).
Since $\alpha R + IJ = I$, $\psi$ is surjective.
It remains to prove that $\psi$ is injective.
Suppose $\alpha x \in IJ$ for $x \in R$.
It suffices to prove that $x \in J$.
Since $\alpha R = IM$, $IMx \subset IJ$.
Since $I$ is invertible, $Mx \subset J$.
Since $M + J = R$, there exist $\mu \in M$ and $\beta \in J$ such that $\mu + \beta = 1$.
Since $\mu x + \beta x = x$, $x \in J$ as desired.