I have the following stochastic differential equation;
\begin{align*} \text{d}S_t &= S_t\mu\text{d}t +\sqrt{V_t}\left(\sqrt{1-\rho^2}\text{d}B_t^{(1)}+\rho\text{d}B_t^{(2)}\right), \quad S_0\geq 0\\ \text{d}V_t &= (\alpha-\lambda V_t)\text{d}t + \sigma_V\sqrt{V_t}\text{d}B_t^{(2)},\quad v_0>0, \quad \lambda,\alpha,\sigma_V\geq 0. \end{align*}
Where $B_t^{(1)},B_t^{(2)}$ are independent Brownian Motions. I need to check that $V$ is independent of $B_t^{(1)}$ and afterwards use this to show that for every function $f$ satisfying $\int_0^Tf(V_s)^2\text{d}s<\infty$, it holds almost surely that
\begin{align}\label{eq:test} \mathbb{E}\left[\mathrm{e}^{\int_0^Tf(V_s)\text{d}B_s^{(1)}}\mathrm{e}^{\int_0^Tf(V_s)^2\text{d}s}\right] = 1 \end{align}
My first problem is that I simply do not know how to show that $V$ and $B_t^{(1)}$ are independent. I can see in the SDE for $V$ that $B_t^{(1)}$ is not present, but i find that argument to be a bit vague.
Secondly for the expectation, I know that $\int_0^Tf(V_s)\text{d}B_s^{(1)}\sim N(0,\int_0^Tf(V_s)^2\text{d}s$). I thought about splitting the expectation into two expectations, but that would require independence, which is something I'm supposed to show earlier. Even though I assume that $V$ and $B_t^{(1)}$ is independent, I still do not know how to show the expectation is equal to 1. Can someone maybe give me an explanation?.
Thanks in advance.