[proofplan] We pass from the original Gaussian coordinates $X_1,\dots,X_n$ to the innovation coordinates $E_1,\dots,E_n$ by the Gram-Schmidt construction in the Hilbert space $L^2(\Omega,\mathcal F,\mathbb P)$. Orthogonality of the innovations makes their covariance matrix diagonal, and Gaussianity turns this diagonal covariance into independence. The coordinate transformation from observations $x$ to innovations $e$ is lower triangular with unit diagonal, so its Jacobian determinant is $1$. The density therefore becomes the product of the one-dimensional Gaussian innovation densities evaluated at $e_t=x_t-\widehat x_t$. [/proofplan]

[step:Construct the innovation coordinates by orthogonal projection] For each $t \in \{1,\dots,n\}$, the space $H_{t-1}$ is finite-dimensional and hence closed in $L^2(\Omega,\mathcal F,\mathbb P)$, so the orthogonal projection $P_{t-1}$ is well-defined. Since $P_{t-1}X_t \in H_{t-1}$, there are real coefficients $a_{tj} \in \mathbb R$, indexed by $1 \le j < t$, such that \begin{align*} \widehat X_t = P_{t-1}X_t = \sum_{j=1}^{t-1} a_{tj}X_j. \end{align*} For $t=1$, this sum is empty and $\widehat X_1=0$. Define the innovation vector \begin{align*} E: \Omega &\to \mathbb R^n \\ \omega &\mapsto (E_1(\omega),\dots,E_n(\omega)). \end{align*} Each $E_t$ is a linear combination of $X_1,\dots,X_t$, namely \begin{align*} E_t = X_t - \sum_{j=1}^{t-1} a_{tj}X_j. \end{align*} Thus $E$ is obtained from $X$ by a lower triangular linear map $B:\mathbb R^n \to \mathbb R^n$ whose diagonal entries are all $1$. [/step]

[step:Prove that the innovations have positive variances] Fix $t \in \{1,\dots,n\}$. Since $E_t=X_t-P_{t-1}X_t$, the defining property of orthogonal projection gives \begin{align*} \mathbb E[E_t Y]=0 \end{align*} for every $Y \in H_{t-1}$. Suppose, toward a contradiction, that $v_t=0$. Since $v_t=\mathbb E[E_t^2]$, this implies $E_t=0$ in $L^2(\Omega,\mathcal F,\mathbb P)$, hence \begin{align*} X_t = \sum_{j=1}^{t-1} a_{tj}X_j \end{align*} in $L^2(\Omega,\mathcal F,\mathbb P)$. Therefore the nonzero vector $c \in \mathbb R^n$ defined by \begin{align*} c_t = 1, \qquad c_j=-a_{tj}\ \text{for }1 \le j<t, \qquad c_j=0\ \text{for }j>t \end{align*} satisfies \begin{align*} c^\top X = 0 \end{align*} in $L^2(\Omega,\mathcal F,\mathbb P)$. Taking second moments gives \begin{align*} 0 = \mathbb E[(c^\top X)^2] = c^\top \Gamma c. \end{align*} This contradicts the nonsingularity of the covariance matrix of a Gaussian vector, which makes $\Gamma$ positive definite. Hence $v_t>0$. [/step]

[step:Use Gaussianity and orthogonality to factor the innovation density]If $s<t$, then $E_s \in H_s \subset H_{t-1}$. Since $E_t$ is orthogonal to $H_{t-1}$, we have \begin{align*} \mathbb E[E_tE_s]=0. \end{align*} Thus the covariance matrix of $E$ is diagonal: \begin{align*} \operatorname{Cov}(E)=\operatorname{diag}(v_1,\dots,v_n). \end{align*} Because $E=BX$ is a linear image of the centred Gaussian vector $X$, the random vector $E$ is centred Gaussian. A centred Gaussian vector with diagonal covariance matrix has independent components, since its characteristic function factors: \begin{align*} \mathbb E[e^{i u\cdot E}] = \exp\left(-\frac12\sum_{t=1}^n v_tu_t^2\right) = \prod_{t=1}^n \exp\left(-\frac12 v_tu_t^2\right) \end{align*} for every $u=(u_1,\dots,u_n)\in \mathbb R^n$. Therefore $E_1,\dots,E_n$ are independent centred Gaussian random variables with variances $v_1,\dots,v_n$. Their joint density with respect to $\mathcal L^n$ is \begin{align*} f_E(e_1,\dots,e_n) = \prod_{t=1}^{n} \frac{1}{(2\pi v_t)^{1/2}} \exp\left(-\frac{e_t^2}{2v_t}\right). \end{align*}[/step]

[guided]The projection construction gives orthogonality, and for Gaussian random variables orthogonality is exactly the same as independence. We verify this directly. Take indices $s<t$. The earlier innovation $E_s$ belongs to $H_s$, because it is a linear combination of $X_1,\dots,X_s$. Since $H_s \subset H_{t-1}$ and $E_t=X_t-P_{t-1}X_t$ is orthogonal to all of $H_{t-1}$, we obtain \begin{align*} \mathbb E[E_tE_s]=0. \end{align*} Therefore distinct innovation coordinates have zero covariance, while the diagonal covariance entries are \begin{align*} \mathbb E[E_t^2]=v_t. \end{align*} Hence \begin{align*} \operatorname{Cov}(E)=\operatorname{diag}(v_1,\dots,v_n). \end{align*} Now use the Gaussian hypothesis. Since each $E_t$ is a linear combination of $X_1,\dots,X_t$, the vector $E$ is a linear image of the Gaussian vector $X$, so $E$ is again centred Gaussian. For a centred Gaussian vector with covariance matrix $\operatorname{diag}(v_1,\dots,v_n)$, the characteristic function is \begin{align*} \mathbb E[e^{i u\cdot E}] = \exp\left(-\frac12 u^\top \operatorname{diag}(v_1,\dots,v_n)u\right) = \exp\left(-\frac12\sum_{t=1}^n v_tu_t^2\right) \end{align*} for every $u=(u_1,\dots,u_n)\in \mathbb R^n$. The final expression factors as \begin{align*} \exp\left(-\frac12\sum_{t=1}^n v_tu_t^2\right) = \prod_{t=1}^n \exp\left(-\frac12 v_tu_t^2\right), \end{align*} which is the product of the one-dimensional characteristic functions of centred Gaussian random variables with variances $v_t$. Therefore $E_1,\dots,E_n$ are independent and have joint density \begin{align*} f_E(e_1,\dots,e_n) = \prod_{t=1}^{n} \frac{1}{(2\pi v_t)^{1/2}} \exp\left(-\frac{e_t^2}{2v_t}\right). \end{align*}[/guided]

[step:Transform from innovation coordinates back to observed coordinates] Define the coordinate map \begin{align*} T:\mathbb R^n &\to \mathbb R^n \\ x &\mapsto e \end{align*} by \begin{align*} e_t = x_t-\sum_{j=1}^{t-1}a_{tj}x_j \end{align*} for $t \in \{1,\dots,n\}$. This is the deterministic version of the random transformation $E=BX$. The matrix of $T$ is lower triangular with diagonal entries equal to $1$, so \begin{align*} |\det T|=1. \end{align*} The change-of-variables formula for Lebesgue density under the invertible linear map $T$ gives \begin{align*} f_X(x)=f_E(Tx)|\det T|. \end{align*} Since $|\det T|=1$ and $(Tx)_t=x_t-\widehat x_t$, we obtain \begin{align*} f_X(x_1,\dots,x_n) = \prod_{t=1}^{n} \frac{1}{(2\pi v_t)^{1/2}} \exp\left(-\frac{(x_t-\widehat x_t)^2}{2v_t}\right). \end{align*} This is the claimed likelihood factorization. [/step]