The mode-$n$ product of a tensor $\mathcal{X}=[x_{i_1,\ldots,i_M}]\in \mathbb{R}^{I_1\times \cdots \times I_M}$ and a matrix $\mathbf{U}=[u_{i_m,j}]\in \mathbb{R}^{I_m\times J}$ is denoted by $\mathcal{X}\times_{m}\mathbf{U}^T$ and is of size $I_1\times \cdots \times I_{m-1}\times J\times I_{m+1}\times \cdots \times I_{M}$. By element, we have $(\mathcal{X}\times_{m}\mathbf{U}^T)_{i_1,\cdots,i_{m-1},j,i_{m+1},\ldots,i_{M}}=\sum_{i_m=1}^{I_m}x_{i_1,\ldots,i_M}u_{i_m,j}$. Let $\mathcal{X}\in \mathbb{R}^{I_1\times I_2\times \cdots \times_{M}I_M}$ and $\mathbf{U}_m\in \mathbb{R}^{I_m\times J_m}$ for all $m \in \lbrace 1,\ldots,M\rbrace$.
The mode-$m$ fiber of a tensor $\mathcal{X}=[x_{i_1,\ldots,i_M}]\in \mathbb{R}^{I_1\cdots\times I_{M}}$ is obtained by fixing every index but $i_m$. The mode-$m$ matricization, is denoted by $\mathbf{X}_{(m)}$ and arranges the mode-$m$ fibers to be the columns of the resulting $I_m\times (I_1\times \cdots \times I_{m-1}\times I_{m+1}\times \cdots \times I_M)$ matrix.
For any $m\in \lbrace 1,\ldots,M \rbrace$, we have the following important property: $$ \begin{array}{lll} \mathcal{Y}&=\mathcal{X}\times_{1}\mathbf{U}_{1}^T\times \cdots \times_{M}\mathbf{U}_{M}^T&\Leftrightarrow \\ & & \mathbf{Y}_{(m)}=\mathbf{U}_{m}^T\mathbf{X}_{(m)}(\mathbf{U}_{M}^T\otimes \cdots\mathbf{U}_{m+1}^T\otimes \mathbf{U}_{m-1}^T\otimes\cdots\otimes \mathbf{U}_{1}^T)^T \end{array} $$ For the sake of simplicity, we denote $\mathcal{X}\times _1 \mathbf{U}_1^T\times \ldots \times_N \mathbf{U}_N^T$ as $\mathcal{X}\prod_{k=1}^{M}\times_{k}\mathbf{U}_{k}^T$ and denote $\mathbf{U}_{M}^T\otimes \cdots\mathbf{U}_{m+1}^T\otimes \mathbf{U}_{m-1}^T\otimes\cdots\otimes \mathbf{U}_{1}^T$ as $\bigotimes_{k=M,k\neq m}^{1}\mathbf{U}_{k}^T$.
Could anyone prove the following fomulation? $$ \mathbf{X}_{(m)}\left(\bigotimes_{k=M,k\neq m}^{1}\mathbf{U}_{k}^T\right)^T=\left(\mathcal{X}\prod_{k=1,k\neq m}^{M}\times_k \mathbf{U}_k^T\right)_{(m)} $$