Let $M_1=X_1+1$ and let $M_n=(\log(M_{n-1})+1)^{X_{n}}$ where $(X_n)$ is an i.i.d. sequence of Poisson(1) distributed random variables. Also, let $(\mathbb{F}_n)$ be the filtration defined by $\mathbb{F_n}=\sigma(X_1,...,X_n)$.
I can see that $M_n$ is indeed $\mathbb{F}_n$ measurable, but how to prove the martingale property $\mathbb{E}(M_n|\mathbb{F_{n-1}})=M_{n-1}$?
Let $X\sim\operatorname{Poi}(\lambda)$ be a Poisson rv with expectation $\lambda$, let $s\in\mathbb R$. Then $$ Es^X = \sum_{k\ge0} s^k \frac{\lambda^k }{k!} e^{-\lambda} = \sum_{k\ge0} \frac{(s\lambda)^k }{k!} e^{-\lambda} = e^{s\lambda}e^{-\lambda} = e^{\lambda(s-1)}.$$ Use this with $X=X_n$, $\lambda=1$ and $s=\log M_{n-1}+1$ to get the desired martingale expectation formula $E(M_n|\mathbb F_{n-1}) = M_{n-1}$.