We know the most popular function of the geometric Brownian motion
$$ \frac{dS_t}{S_t} = \mu dt + \sigma dB_t $$
We can write the geometric Brownian motion as the discretized version
$$ \frac{S_n - S_{n-1}}{S_{n-1}} = \mu\Delta + \sigma \sqrt{\Delta} Z_n $$
If we have a fixed time $T > 0$, divide $[0,T]$ into $N$ equal segments, $\Delta = \frac{T}{N}$. Moreover, based on the solution of geometric Brownian motion, we can write in another form such that $\nu = \mu - \frac{\sigma^2}{2}$,
$$ S_n = S_{n-1}e^{\nu\Delta + \sigma \sqrt{\Delta} Z_n} $$
Could someone provide proof of why these two functions are equivalent when $\Delta$ tends to zero? Thank you very much.