On a fixed interval $[0,T]$, let $(W_t)_{0\le t \le T}$ be a Brownian motion, and $ (\gamma_t)_{0\le t \le T} $ a cadlag process. Let $$ M_t = exp ({\int_0^t\gamma_sdW_s - \frac{1}{2}\int_0^t\gamma_s^2ds}) \\$$ for $0 \le t \le T$
Show that $M_t$ is a martingale.
I started by applying the definition of martingale, which is $$E[M_s|\Im_t]=E[M_s\frac{M_t}{M_s}|\Im_t]$$ so$$E[M_s|\Im_t]=M_sE[\frac{M_t}{M_s}|\Im_t]$$ Then I tried to substitute $M_t = exp ({\int_0^t\gamma_sdW_s - \frac{1}{2}\int_0^t\gamma_s^2ds}) $
and $M_s = exp ({\int_0^s\gamma_sdW_s - \frac{1}{2}\int_0^s\gamma_s^2ds})$ into $E[\frac{M_t}{M_s}|\Im_t]$.
I am not sure i am heading on the right track and also after I did substitution I dont know how to use cadlag process properties to solve this problem.
If you are familiar with Itô's lemma, then here is a hint:
Use Itô's lemma to show that $$dM_t = - \gamma_t M_t dW_t.$$ The above immediately implies that $M_t$ is a (positive) martingale.