A Kronecker Product identity I want to show the Kronecker Product identity listed on Wikipedia:
$$\begin{align}
\mathrm{vec}(AXB)
=(B^T \otimes A) \mathrm{vec}(X) \\ \tag{1}
\end{align}$$
Wikipedia does not cite references for the proof for (1), and I want to verify.
Before the calculation, agree with some convention.
Let us denote $a\; \mathrm{mod}\; b$ for the remainder of $a$ modulo $b$ (so that $9\; \mathrm{mod}\; 4 =1$).
This is unlike the usual practice, but otherwise the expression gets too verbose.
Moreover let us denote the $(i,j)$-th entry of $A$ to be $A_{i,j}$, and $i$-th entry of $v$ to be $v_{i}$.
Also say that $A \in \mathbb{R}^{M \cdot N}$ if it is a $M$ by $N$ matrix, by slightly abusing the meaning of $\mathbb{R}$.
If we stick to only addition and multiplication, once the proof works, it works for any field.
Firstly, restate the formal definition of Kronecker product and vectorization, for later use.
For $X' \in \mathbb{R}^{N_2 \cdot N_3}$
$$\begin{align}
(\mathrm{vec}(X'))_{m_1}
=X'_{m_1\; \mathrm{mod}\; N_1, \lceil m_1/N_1 \rceil} \\ \tag{2}
\end{align}$$
For $A' \in \mathbb{R}^{N_1 \cdot N_2}, B' \in \mathbb{R}^{N_3 \cdot N_4}$,
$$\begin{align}
(A' \otimes B')_{m_1, m_2}
=A'_{\lceil m_1/N_3 \rceil, \lceil m_2/N_4 \rceil}
B'_{m_1\; \mathrm{mod}\; N_3, m_2\; \mathrm{mod}\; N_4} \\ \tag{3}
\end{align}$$ 
Now, let $A \in \mathbb{R}^{N_1 \cdot N_2},\; X \in \mathbb{R}^{N_2 \cdot N_3},\; B \in \mathbb{R}^{N_3 \cdot N_4},$,
We try to evaluate lhs and rhs of (1).
$$\begin{align}
&(\mathrm{vec}(AXB))_{m_1} \\
=&(AXB)_{m_1\; \mathrm{mod}\; N_1, \lceil m_1/N_1 \rceil} \\
=&\sum_{n_2=1}^{N_2} \sum_{n_3=1}^{N_3}
A_{m_1\; \mathrm{mod}\; N_1, n_2} X_{n_2, n_3} B_{n_3, \lceil m_1/N_1 \rceil} \\ \tag{4}
\end{align}$$
$$\begin{align}
&((B^T \otimes A) \mathrm{vec}(X) )_{m_1} \\
=&\sum_{m_2=1}^{N_2 N_3} (B^T \otimes A)_{m_1, m_2} \mathrm{vec}(X)_{m_2} \\
=&\sum_{m_2=1}^{N_2 N_3} (B^T)_{\lceil m_1/N_1 \rceil, \lceil m_2/N_2 \rceil}
A_{m_1\; \mathrm{mod}\; N_1, m_2\; \mathrm{mod}\; N_2} \mathrm{vec}(X)_{m_2} \\ 
=&\sum_{m_2=1}^{N_2 N_3} B_{\lceil m_2/N_2 \rceil, \lceil m_1/N_1 \rceil}
A_{m_1\; \mathrm{mod}\; N_1, m_2\; \mathrm{mod}\; N_2} X_{m_2\; \mathrm{mod}\; N_2, \lceil m_2/N_2 \rceil} \\ \tag{5}
\end{align}$$ 
It would seem that we are close, but it is not obvious that (4) and (5) are equal.
Although $A$'s row index $m_1\; \mathrm{mod}\; N_1$ and $B$'s column index $\lceil m_1/N_1 \rceil$ agree, the other do not.
It is suggestive that maybe if we "sum away" $n_2, n_3$, we might find them to agree, but $A,X,B$ are all free, and we have nothing to exploit to carry out that summation.
 A: Your effort is valiant, but I can tell you from experience that it is helpful to avoid getting into the quagmire of modular arithmetic (or of directly indexing of a Kronecker product at all) when possible.  Here's how I would prove the statement:
First, it helps to note that for any column-vectors $u,v,$ we have 
$$
\operatorname{vec}(uv^T) = v \otimes u
$$
I think you will find proving this statement to be a useful exercise. In fact, I prefer to take the above equation to be the definition of the vectorization operator (wikipedia defines the operator as "stacking the columns"). 
Let $e_1,\dots,e_n$ denote the canonical basis (i.e. the columns of the $n \times n$ identity matrix).  Let $X = (x_{ij})_{i,j = 1}^{m,n}$. We have
$$
\begin{align}
\operatorname{vec}(AXB) &= 
\operatorname{vec}\left(A\left(\sum_{i=1}^m \sum_{j=1}^n x_{ij} e_ie_j^T\right)B\right) 
\\ & =
\sum_{i=1}^m\sum_{j=1}^n x_{ij} \operatorname{vec}(Ae_ie_j^TB)
\\ & = \sum_{i=1}^m\sum_{j=1}^nx_{ij} \operatorname{vec}([Ae_i][B^Te_j]^T) 
\\ &= \sum_{i=1}^m\sum_{j=1}^nx_{ij} [B^Te_j]\otimes [Ae_i]
\\ &= (B^T \otimes A)\left(\sum_{i=1}^m\sum_{j=1}^nx_{ij} e_j \otimes e_i\right)
\\ &= (B^T \otimes A)\left(\sum_{i=1}^m\sum_{j=1}^nx_{ij} \operatorname{vec}(e_i e_j^T)\right)
\\ &= (B^T \otimes A)\operatorname{vec}\left(\sum_{i=1}^m\sum_{j=1}^nx_{ij} e_i e_j^T\right) = (B^T \otimes A)\operatorname{vec}(X)
\end{align}
$$
A: I see! In (5), as $m_2 =1, \dots, N_2 N_3$, we see that $m_2\; \mathrm{mod}\; N_2 =1, \dots N_2$, and $\lceil m_2/N_2 \rceil =1, \dots N_3$, same as the numbering of $n_2, n_3$ respectively in (4).
