Transpose of block matrix

Most people would just claim this is obvious and omit the proof, but if you don't want to do that then perhaps you could first prove that \begin{equation} \begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix} \end{equation} and \begin{equation} \begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}. \end{equation} Then \begin{align} \begin{bmatrix} A & B \\ C & D \end{bmatrix}^T &= \begin{bmatrix} \begin{bmatrix} A \\ C \end{bmatrix}^T \\ \begin{bmatrix} B \\ D \end{bmatrix}^T \end{bmatrix} \\ &= \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix}. \end{align}