Let $\{M_n\}_{n\ge 0}$ be a symmetric simple random walk. Fix a real $b$. Prove that the process $S_n = e^{bM_n} (\frac{2}{e^b + e^{-b}})^n$, $n = 0,1,2,....$, is a martingale w.r.t. the natural filtration $\{\mathcal{F}_n\}_{n\ge 0}$.
I know that a stochastic process is called a martingale if $E[M_{t+s} | \mathcal{F}] = M_t$ but I am confused about how to start this proof. Any help or hints are greatly appreciated!!
Notice that $M_n$ is $\mathcal F_n$ measurable hence so is $S_n$, and $M_n$ is integrable since it is bounded.
It remains to check that $\mathbb E(S_{n+1} \mid \mathcal F_n)=S_n$. To this aim, we have to compute $\mathbb E(e^{bM_{n+1}} \mid \mathcal F_n)$. As $$e^{bM_{n+1}}=\underbrace{e^{bM_{n+1}-bM_n}}_{\mbox{independent of }\mathcal F_n}\cdot \underbrace{e^{bM_{n}}}_{\mathcal F_n\mbox{-measurable}},$$ we get $$\mathbb E(e^{bM_{n+1}} \mid \mathcal F_n)=e^{bM_{n}}\mathbb E(e^{bM_{n+1}-bM_n}).$$