I am having some difficulty proving the following statement taken from Oksendals SDE's exercise 2.11.
If $B_{t} = (B^{(1)}_{t},...,B^{(n)}_t)$ is an $n$-dimensional Brownian motion, then the $1$-dimensional processes $\{B^{(j)}_t\}_{t \geq 0}, \ 1 \leq j \leq n $ are independent, $1$-dimensional Brownian motions.
my attempt
I originally began by trying to prove that
$E[B^{(1)}_{t}...B^{(n)}_{t}]=E[B^{(1)}_{t}]...E[B^{(n)}_{t}]$
The RHS is straightforward as $E[B^{(j)}_{t}] = 0 $ since $B^{(j)}_{t}$ has distribution $\mathcal{N}(0,t)$ for $1 \leq j \leq n$.
However I am stuck on showing the LHS is $E[B^{(1)}_{t}...B^{(n)}_{t}] = 0$
My other thought was to try prove that the covariance matrix ($C$) of $B_{t}$ is $C = t\mathbb{I}$ where $\mathbb{I}$ is the identity matrix. But I come across the same issue trying to show that $E[B^{(i)}_{t}B^{(j)}_{t}]=0$ for $i \neq j$