Deriving the maximum likelihood for the parameters in linear regression Notation:
$\textbf{w}$ is an M-dimensional vector of parameters (including the bias parameter), $\textbf{x}_n$ is an M-dimensional vector of the features of each training example, $\textbf{t}$ is an N-dimensional vector of target values - where $t_n$ is the output of the nth training example and finally, $\Phi$ is an $M \times N$ design matrix.
During studying linear regression, I naturally came across the derivation of the normal equation. In his book, Christopher M. Bishop begins by the log of the likelihood function given by:
$$
\frac{N}{2} \ln \beta - \frac{N}{2} \ln(2\pi) - \frac{\beta}{2}\sum_{n=1}^N\{t_n - \textbf{w}^T \phi(\textbf{x}_n)\}^2 \tag{1}
$$
With the goal of calculating the maximum likelihood of the parameters, he then calculates the derivative with respect to $w$:
$$
\bigtriangledown p(\textbf{t}| \textbf{w},\beta) = \beta \sum_{n=1}^N\{t_n - \textbf{w}^T \phi(\textbf{x}_n)\}\phi(\textbf{x}_n)^T \tag 2
$$
After which he sets the derivative to zero, in order to maximise the likelihood function
$$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
So far so good. But then, the result for the maximum likelihood of the parameters $w$ is written as:
$$
\textbf{w}_{ML} = (\Phi^T\Phi)^{-1} \Phi^T \textbf{t} \tag 4
$$
My own derivation from equation $(3)$ to equation $(4)$ goes along the following lines:
\begin{align*}
\Phi^T \Phi \textbf{w} &= \Phi^T \textbf{t} \tag 5 \\ 
(\Phi^T \Phi)^{-1} \Phi^T \Phi \textbf{w} &= (\Phi^T \Phi)^{-1} \Phi^T \textbf{t} \tag 6 \\
\textbf{w} &= (\Phi^T \Phi)^{-1} \Phi^T \textbf{t} \tag 7
\end{align*}
My question is this: How does one go from equation $(3)$ to equation $(4)$. How does the supposed $ \textbf{w}^T \phi(\textbf{x}_n) \phi(\textbf{x}_n)$ become $\Phi^T \Phi \textbf{w}$ ? I do of course understand that given the dimensions of the matrix $\Phi$ and the vector $\textbf{w}$, that this is the only way to multiply the two entities. But what is the steps that one would follow in a formal derivation in order to reach such a result?
 A: $$
0 = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T - \textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) \tag 3
$$
$$
\textbf{w}^T \bigg(\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T \bigg) = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T  \tag a
$$

Recall the Delta Matrix is defined as: $$ \Phi =\begin{pmatrix} \phi _{ 0 }(x_{ 1 }) & \phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ M-1 }(x_{ 1 }) \\ \phi _{ 0 }(x_{ 2 }) & \phi _{ 1 }(x_{ 2 }) & \dots  & \phi _{ M-1 }(x_{ 2 }) \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \phi _{ 0 }(x_{ N }) & \phi _{ 1 }(x_{ N }) & \dots  & \phi _{ M-1 }(x_{ N }) \end{pmatrix}$$


We can show that below
$$ \sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T = \Phi^T\Phi$$

$$
\begin{equation}
\begin{split}
\sum_{n=1}^N \phi(\textbf{x}_n)\phi(\textbf{x}_n)^T  & = \begin{bmatrix} \phi _{ 0 }(x_{ 1 }) \\ \phi _{ 1 }(x_{ 1 }) \\ \vdots  \\ \phi _{ M-1 }(x_{ 1 }) \end{bmatrix}\begin{bmatrix} \phi _{ 0 }(x_{ 1 }) & \phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ M-1 }(x_{ 1 }) \end{bmatrix}+\dots \\
 & +\begin{bmatrix} \phi _{ 0 }(x_{ N }) \\ \phi _{ 1 }(x_{ N }) \\ \vdots  \\ \phi _{ M-1 }(x_{ N }) \end{bmatrix}\begin{bmatrix} \phi _{ 0 }(x_{ N }) & \phi _{ 1 }(x_{ N }) & \dots  & \phi _{ M-1 }(x_{ N }) \end{bmatrix}\\
& = \begin{bmatrix} \phi _{ 0 }(x_{ 1 })\phi _{ 0 }(x_{ 1 }) & \phi _{ 0 }(x_{ 1 })\phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ 0 }(x_{ 1 })\phi _{ M-1 }(x_{ 1 }) \\ \phi _{ 1 }(x_{ 1 })\phi _{ 0 }(x_{ 1 }) & \phi _{ 1 }(x_{ 1 })\phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ 1 }(x_{ 1 })\phi _{ M-1 }(x_{ 1 }) \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \phi _{ M-1 }(x_{ 1 })\phi _{ 0 }(x_{ 1 }) & \phi _{ M-1 }(x_{ 1 })\phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ M-1 }(x_{ 1 })\phi _{ M-1 }(x_{ 1 }) \end{bmatrix}+\dots \\
& +\begin{bmatrix} \phi _{ 0 }(x_{ N })\phi _{ 0 }(x_{ N }) & \phi _{ 0 }(x_{ N })\phi _{ 1 }(x_{ N }) & \dots  & \phi _{ 0 }(x_{ N })\phi _{ M-1 }(x_{ N }) \\ \phi _{ 1 }(x_{ N })\phi _{ 0 }(x_{ N }) & \phi _{ 1 }(x_{ N })\phi _{ 1 }(x_{ N }) & \dots  & \phi _{ 1 }(x_{ N })\phi _{ M-1 }(x_{ N }) \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \phi _{ M-1 }(x_{ N })\phi _{ 0 }(x_{ N }) & \phi _{ M-1 }(x_{ N })\phi _{ 1 }(x_{ N }) & \dots  & \phi _{ M-1 }(x_{ N })\phi _{ M-1 }(x_{ N }) \end{bmatrix} \\
& = \begin{bmatrix} \sum _{ i }^{ N }{ \phi _{ 0 }(x_{ i })\phi _{ 0 }(x_{ i }) }  & \sum _{ i=1 }^{ N }{ \phi _{ 0 }(x_{ i })\phi _{ 1 }(x_{ i }) }  & \dots  & \sum _{ i=1 }^{ N }{ \phi _{ 0 }(x_{ i })\phi _{ M-1 }(x_{ i }) }  \\ \sum _{ i=1 }^{ N }{ \phi _{ 1 }(x_{ i })\phi _{ 0 }(x_{ i }) }  & \sum _{ i=1 }^{ N }{ \phi _{ 1 }(x_{ i })\phi _{ 1 }(x_{ i }) }  & \dots  & \sum _{ i=1 }^{ N }{ \phi _{ 1 }(x_{ i })\phi _{ M-1 }(x_{ i }) }  \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \sum _{ i=1 }^{ N }{ \phi _{ M-1 }(x_{ i })\phi _{ 0 }(x_{ i }) }  & \sum _{ i=1 }^{ N }{ \phi _{ M-1 }(x_{ i })\phi _{ 1 }(x_{ i }) }  & \dots  & \sum _{ i=1 }^{ N }{ \phi _{ M-1 }(x_{ i })\phi _{ M-1 }(x_{ i }) }  \end{bmatrix} \\ 
& = \begin{bmatrix} \phi _{ 0 }(x_{ 1 }) & \phi _{ 0 }(x_{ 2 }) & \dots  & \phi _{ 0 }(x_{ N }) \\ \phi _{ 1 }(x_{ 1 }) & \phi _{ 1 }(x_{ 2 }) & \dots  & \phi _{ 1 }(x_{ N }) \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \phi _{ M-1 }(x_{ 1 }) & \phi _{ M-1 }(x_{ 2 }) & \dots  & \phi _{ M-1 }(x_{ N }) \end{bmatrix}\begin{bmatrix} \phi _{ 0 }(x_{ 1 }) & \phi _{ 1 }(x_{ 1 }) & \dots  & \phi _{ M-1 }(x_{ 1 }) \\ \phi _{ 0 }(x_{ 2 }) & \phi _{ 1 }(x_{ 2 }) & \dots  & \phi _{ M-1 }(x_{ 2 }) \\ \vdots  & \vdots  & \ddots  & \vdots  \\ \phi _{ 0 }(x_{ N }) & \phi _{ 1 }(x_{ N }) & \dots  & \phi _{ M-1 }(x_{ N }) \end{bmatrix} \\ 
& = \Phi^T\Phi
\end{split}
\end{equation}$$
Therefore we get the following.
$$
\textbf{w}^T \bigg(\Phi^T\Phi \bigg) = \sum_{n=1}^N t_n \phi(\textbf{x}_n)^T  \tag b
$$
Next we transpose both sides
$$
\bigg( \textbf{w}^T \bigg(\Phi^T\Phi \bigg) \bigg)^T = \bigg(\sum_{n=1}^N t_n \phi(\textbf{x}_n)^T \bigg)^T  \tag b
$$
$$
\Phi^T\Phi \textbf{w} = \sum_{n=1}^N t_n \phi(\textbf{x}_n)  \tag c
$$
Similarly, $$\sum_{n=1}^N t_n \phi(\textbf{x}_n) = t_{ 1 }\begin{bmatrix} \phi _{ 0 }(x_{ 1 }) \\ \phi _{ 1 }(x_{ 1 }) \\ \vdots  \\ \phi _{ M-1 }(x_{ 1 }) \end{bmatrix}\quad +\quad \dots \quad +\quad t_{ N }\begin{bmatrix} \phi _{ 0 }(x_{ N }) \\ \phi _{ 1 }(x_{ N }) \\ \vdots  \\ \phi _{ M-1 }(x_{ N }) \end{bmatrix}\quad =\quad \Phi ^{ T }{ t }$$
Finally we get 
$$
\Phi^T\Phi \textbf{w} = \Phi ^{ T }{ t }  \tag 5
$$
