Let $X_0^i$ be iid random variables with law $\mu_0$ for $i=1,\dots, N$. Let $\{X_t^i\}_{1\le i\le N}$ be satisfied as the following OU process: for $i=1,\dots, N$: $$ dX_t^i=-X_t^idt+\sqrt{2}dB_t^i. $$
Why can we say $$ \exp\left(\sum_{i=1}^N\left(\int_0^t X_s^i dB_s^i-\frac{1}{2}\int_0^t (X_s^i)^2ds\right)\right) $$ is a martingale?
It seems that it follows from the result that
[textbook by Le Gall] for $Z_t=\int_0^t f(s)dB_s$ where $f(s)\in L^2$, then $$ \exp(Z_t-\frac{1}{2}[Z]_t)=\exp(\int_0^t f(s)dB_s-\frac{1}{2}\int_0^t f^2(s)ds) $$ is a martingale.
But I have no idea why the first display is true because that is to take $$ Z_t=\sum_{i=1}^N\int_0^t X_t^i dB_t^i. $$
By https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process or Ito formula, we can get $$ X_t^i=X_0^ie^{-t}+\sqrt{2}\int_0^t e^{-(t-s)}dB_s^i. $$