Proving that ${\rm vec}(A\,{\rm Diag}(b)\,C) = ((C^T\otimes 1_a)\odot(1_c\otimes A))\,b$ Given the following vectors and matrices 
$$\eqalign{
 &A\in{\mathbb R}^{a\times b},\,\,\,\,
 &B\in{\mathbb R}^{b\times b},\,\,\,\,
 &C\in{\mathbb R}^{b\times c} \cr
 &1_a\in{\mathbb R}^{a\times 1},\,\,\,\,
 &b\in{\mathbb R}^{b\times 1},\,\,\,\,
 &1_c\in{\mathbb R}^{c\times 1} \cr
}$$
where $B={\rm Diag}(b)\,$ and $\,1_n$ denotes a vector of all ones of length $n$.
I would like to show that the vector $\,v={\rm vec}(ABC)\,$ can be expanded as 
$$\eqalign{
v &= \Big((C^T\otimes 1_a)\odot(1_c\otimes A)\Big)\,b
}$$
where $(\otimes, \odot)$ denote the Kronecker and Hadamard products, respectively.
I am aware of several other expressions for this vector 
$$\eqalign{
 v &= (C^T\otimes A)\,{\rm vec}(B) \cr
 v &= \Big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\Big)\,{\rm vec}(B) \cr
}$$
but I don't see how to arrive at the desired formula.


Update
After studying Omnomnomnom's answer, I realized that I needed to exploit several esoteric properties to prove the formula.
1) The outer product of two vectors vectorizes to their Kronecker product
$$\eqalign{
 {\rm vec}(ab^T) &= b\otimes a \cr
}$$
2) Vectors from the canonical basis distribute over a Hadamard product
$$\eqalign{
 (M\odot N)e_k &= (Me_k)\odot(Ne_k) \cr
}$$
3) The distribution property of the Kronecker product of 2 arbitrary vectors and a matrix
$$\eqalign{
 (C^T\otimes 1)e &= {\rm vec}(1(e^TC)) = (C^Te)\otimes 1 \cr
 (1\otimes A)e &= {\rm vec}((Ae)1^T) = 1\otimes(Ae) \cr
}$$
4) A rule for mixed Kronecker/Hadamard products 
$$\eqalign{
 (M\odot N)\otimes(P\odot Q) &= (M\otimes P)\odot(N\otimes Q) \cr\cr
}$$
Use these rules to evaluate the $k^{th}$ column of the anticipated solution
$$\eqalign{
&\big((C^T\otimes 1_a)\odot(1_c\otimes A)\big)\,e_k \cr
&(C^T\otimes 1_a)e_k\odot(1_c\otimes A)e_k \cr
&(C^Te_k\otimes 1_a)\odot(1_c\otimes Ae_k) \cr
&(C^Te_k\odot 1_c)\otimes(1_a\odot Ae_k) \cr
&(C^Te_k)\otimes(Ae_k) \cr
}$$
which matches the $k^{th}$ column of Omnomnomnom's matrix.


Update #2
I also like O's second approach, which I interpret as
$$\eqalign{
&\Big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\Big)\,{\rm vec}(B)\cr
&=\sum_k b_k\,\,\big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\big)\,\big(e_k\otimes e_k\big) \cr
&=\sum_k b_k\,\,\big((C^T\otimes 1_a1_b^T)(e_k\otimes e_k)\big)\odot\big((1_c1_b^T\otimes A)(e_k\otimes e_k)\big) \cr
&=\sum_k b_k\,\,\big(C^Te_k\otimes 1_a1_b^Te_k\big)\odot\big(1_c1_b^Te_k\otimes Ae_k\big) \cr
&=\sum_k b_k\,\,\big(C^Te_k\otimes 1_a\big)\odot\big(1_c\otimes Ae_k\big) \cr
&=\sum_k b_k\,\,\big((C^T\otimes 1_c)\odot(1_a\otimes A)\big)e_k \cr
&=\big((C^T\otimes 1_c)\odot(1_a\otimes A)\big)\,b \cr
}$$
 A: Let $e_1,e_2,\dots,e_b$ denote the canonical basis of $\Bbb R^b$.  We compute
$$
\begin{align*}
\operatorname{vec}(A \operatorname{diag}(b) C) &= 
\operatorname{vec}\left(A \left[\sum_{i=1}^b b_i e_ie_i^T\right] C\right)
\\ & = 
\operatorname{vec}\left(\sum_{i=1}^b b_i (Ae_i)(C^Te_i)^T\right)
\\ & =
\sum_{i=1}^b b_i \,(C^Te_i) \otimes (Ae_i)
\\ & = 
\pmatrix{(C^Te_1) \otimes (Ae_1) & \cdots & (C^Te_b) \otimes (Ae_b)} \pmatrix{b_1\\ \vdots \\ b_b}
\end{align*}
$$
(as you did, I have used $b$ to indicate both the diagonal vector of $B$ and the size of $B$).
It then suffices to rewrite the matrix on the left as
$$
\pmatrix{(C^Te_1) \otimes (Ae_1) & \cdots & (C^Te_b) \otimes (Ae_b)} = 
\Big((C^T\otimes 1_a)\odot(1_c\otimes A)\Big)
$$
One way to do so is to write
$$
\pmatrix{(C^Te_1) \otimes (Ae_1) & \cdots & (C^Te_b) \otimes (Ae_b)} =\\
\pmatrix{\operatorname{vec}([Ae_1][C^Te_1]^T) & \cdots & \operatorname{vec}([Ae_b][C^Te_b]^T)}
$$
and from there, apply your Hadamard formula to each column to see that the $i$th column is indeed $((C^Te_i) \otimes 1_a) \odot (1_c \otimes (Ae_i))$, so that the matrix on the left is indeed $(C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)$.

Another approach: it suffices to check that
$$
\Big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\Big)\,{\rm vec}(e_ie_i^T) = 
((C^Te_i) \otimes 1_a) \odot (1_c \otimes (Ae_i))
$$
Since we have
$$
\Big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\Big)\,{\rm vec}(B) = 
\sum_{i=1}^b b_i\,\Big((C^T\otimes 1_a1_b^T)\odot(1_c1_b^T\otimes A)\Big){\rm vec}(e_ie_i^T)
$$

Regarding the end of my first approach: I meant that we could use your last formula to note that 
$$
\operatorname{vec}([Ae_k][C^Te_k]^T) = \operatorname{vec}([Ae_k]_{a \times 1}\,[1]_{1 \times 1}\,[C^Te_k]^T_{1 \times c}) =\\
\Big(([C^Te_k]\otimes 1_a1_1^T)\odot(1_c1_1^T\otimes [Ae_k])\Big)\,{\rm vec}([1]_{1 \times 1}) = \\
([C^Te_k] \otimes 1_a) \odot (1_c \otimes [Ae_k])
$$
