Proving that $ (\hat{\beta} - \beta)' (X' X) (\hat{\beta} - \beta)$ is independent with SSE Exercise:

Prove that $ \mathbf{(\hat{\beta} - \beta)' (X' X) (\hat{\beta} - \beta)}$ and SSE are independent for a Least Squares Regression Model.

Attempt:
Note that by $'$ I denote the transpose matrix of the given quantity. I will also use the notation $ \beta: = b$ for the sake of simplification to the code.
I tried, first of all, to yield a matrix expression for SSE so that it can be useful to make conclusions and calculations with the matrix expressions given :
\begin{align*}
\text{SSE} &= \mathbf{e'e = (y-X\hat{b})'(y-X\hat{b})}\\
           &= \mathbf{y'y - 2\hat{b}'X'y + \hat{b}'X'X\hat{b}}\\
           &= \mathbf{y'y - 2\hat{b}'X'y + \hat{b}'X'X(X'X)^{-1}X'y}\\
           &= \mathbf{y'y - 2\hat{b}'X'y+\hat{b}'IX'y}\\
           &= \mathbf{y'y - \hat{b}'X'y}\\
\end{align*}
I assume that now, to show the independance, I need to show that the Covariance of them is equal to zero, namingly :
$$\text{Cov}\Big(\mathbf{(\hat{b} - b)' (X' X) (\hat{b} - b), \mathbf{y'y - \hat{b}'X'y}}\Big)$$
$$=$$
$$\mathbb{E}\Big(\mathbf{(\hat{b} - b)' (X' X) (\hat{b} - b)\big(\mathbf{y'y - \hat{b}'X'y}\big)\Big)}$$
$$-\mathbb{E}\Big(\mathbf{(\hat{b} - b)' (X' X) (\hat{b} - b)}\Big)\mathbb{E}\Big(\mathbf{y'y - \hat{b}'X'y}\Big)$$
But this seems very, very complicated and I cannot see how to continue. I don't even know if this approach is correct or if there exists a simpler way to show the independance between the two quantities given.
Any hints or even better thorough elaborations will be very much appreciated.
 A: Since $[X(\hat\beta-\beta)]^TX(\hat\beta-\beta)$ is a function of $X(\hat\beta-\beta)$ and $\operatorname{SSE}:=(Y-X\hat\beta)^T(Y-X\hat\beta)$ is a function of $Y-X\hat\beta$, to prove independence it's enough to show that the random vectors
$Y-X\hat\beta$ and $X(\hat\beta-\beta)$ are jointly Gaussian and have zero covariance.
Introduce the hat matrix
$
H:=X(X^TX)^{-1}X^T$. Check that $H^2=H$ and $HX=X$ and $HY=X\hat\beta$, where $\hat\beta:=(X^TX)^{-1}X^TY$ is the vector of estimated parameters. These facts imply that
$$X\hat\beta=HY=H(X\beta+\epsilon)=X\beta+H\epsilon\tag1
$$
and
$$Y-X\hat\beta=Y-HY=(I-H)Y=(I-H)(X\beta +\epsilon)=(I-H)\epsilon.\tag2
$$
From (2) and (1) it's clear that $Y-X\hat\beta$ and $X(\hat\beta-\beta)$ are jointly Gaussian. Moreover, we can easily compute their covariance:
$$\begin{align}
\operatorname{Cov}\big(Y-X\hat\beta,X(\hat\beta-\beta)\big)
&\stackrel{(2,1)}=\operatorname{Cov}\big((I-H)\epsilon,H\epsilon\big)\\
&\stackrel{(3)}=E[(I-H)\epsilon(H\epsilon)^T]\\
&\stackrel{(4)}=(I-H)E(\epsilon\epsilon^T)H^T\\
&\stackrel{(5)}=\sigma^2 (I-H)H\\
&\stackrel{(6)}=0.
\end{align}
$$
In (3) we use the fact that $E(\epsilon)=0$. In (4) we see that everything except $\epsilon\epsilon^ T$ is non-stochastic. In (5) we use $E(\epsilon\epsilon^ T)=\sigma^2I$ and that $H$ is symmetric. In (6) we use the idempotency property $H^2=H$.
Note: The independence of the random vectors $Y-X\hat\beta$ and $X(\hat\beta-\beta)$ is distinct from their (pointwise) orthogonality, which is proved as follows:
$$\big(Y-X\hat\beta\big)^T\big(X(\hat\beta-\beta)\big)=
\big((I-H)\epsilon\big)^ T\big(H\epsilon\big)=\epsilon^T\underbrace{(I-H)H}_0\epsilon
$$
