Pairwise independence implies independence for normal random variables I'm reading a book on Brownian Motion. In the proof of the existence of such random function (Wiener, 1923), the following is stated:

Indeed, all increments $B(d)-B(d-2^{-n})$, for $d\in \mathcal{D}_n\setminus \{0\}$, are independent. To see this it suffices to show that they are pairwise independent. as the vector of these increments is Gaussian.

The last part of this quote is the claim that pairwise independent normal variables from a Gaussian family are independent. Could anyone provide/direct me to a proof of this claim? 
Thanks!
 A: Actually it is rather straight forward:
Assume you have a gaussian vector (so the joint distribution follows a gaussian distribution) 
$$
\mathbf X=(X_1,X_2,\ldots,X_n)\sim N(\mathbf \mu,\Sigma)
$$
where $\Sigma\in M^{n\times n}(\mathbb R)$ describes the covariance matrix. The density function looks like 
$$
f_{\mathbf X}(x_1,\ldots,x_n) =
\frac{1}{\sqrt{(2\pi)^{n}\lvert\Sigma\rvert}}
\exp\left(-\frac{1}{2}({\mathbf X}-{\mathbf\mu})^\mathrm{T}{\Sigma}^{-1}({\mathbf X}-{\mathbf\mu})
\right),
$$
Now, we have independence (mutually) iff the density function factorizes. Since the random variables are pairwise independent (uncorrelated suffices), the covariance matrix is a diagonal matrix
$$
\Sigma=\mathrm{diag}(\sigma_1,\ldots,\sigma_n) \text{ and }\mathbf \mu=(m_1,\ldots,m_n)
$$
since the above holds, the density indeed factorizes and we get
$$
f_{\mathbf X}(x_1,\ldots,x_n)=\prod_{i=1}^n\frac{1}{\sqrt{2\pi{\sigma_i}^2}}\exp\bigg(-\frac{{(x_i-m_i)}^2}{2{\sigma_i}^2}\bigg)
$$
and we have mutually independent gaussians. 
