Showing that a random process is a martingale

Let $(M_n)$ be a $(\mathcal{F}_n)$-martingale with $M_0=0$ and note $D_n = M_{n+1}- M_n$. Let $s>0$ and for $n\ge 0$ :

$$N_n = \exp \left( sM_n-\sum_{i=1}^n \log \mathbb{E}(e^{sD_i} |\mathcal{F}_i) \right) $$

Show that $(N_n)$ is a positive martingale for $(\mathcal{F}_n)$.

My take :

$$\mathbb{E}(N_{n+1}|\mathcal{F}_n) = \mathbb{E} \left( \frac{e^{sM_{n+1}}}{\prod_{i=1}^{n+1}\mathbb{E}(e^{sD_i} |\mathcal{F}_i)} \bigg| \mathcal{F}_n \right) = \frac{\mathbb{E} \left( \frac{e^{sM_{n+1}}}{\mathbb{E}(e^{sD_{n+1}} |\mathcal{F}_{n+1})} \bigg| \mathcal{F}_n \right)}{\prod_{i=1}^{n}\mathbb{E}(e^{sD_i} |\mathcal{F}_i)}$$

Now it would be nice if we could show that : $$\mathbb{E} \left( \frac{e^{sM_{n+1}}}{\mathbb{E}(e^{sD_{n+1}} |\mathcal{F}_{n+1})} \bigg| \mathcal{F}_n \right) = e^{sM_n} $$

but how ?

Write $$ e^{sM_{n+1}}=e^{sM_{n}+sD_n} $$ hence $$ \mathbb{E}(N_{n+1}|\mathcal{F}_n) = N_n \mathbb{E} \left( \frac{e^{sD_{n}}}{\mathbb{E}(e^{sD_{n+1}} |\mathcal{F}_{n+1})} \bigg| \mathcal{F}_n \right) $$ but we may not have $N_n$: suppose that $M_n=\sum_{i=1}^n\xi_i$ where $(\xi_i)$ is independent and centered and let $\mathcal F_n$ be generated by $\xi_1,\dots,\xi_n$. Then $D_n=\xi_{n+1}$ and $$ \mathbb{E} \left( \frac{e^{sD_{n}}}{\mathbb{E}(e^{sD_{n+1}} |\mathcal{F}_{n+1})} \bigg| \mathcal{F}_n \right)=\mathbb{E} \left( \frac{e^{s\xi_{n+1}}}{\mathbb{E}(e^{s\xi_{n+2}} |\mathcal{F}_{n+1})} \bigg| \mathcal{F}_n \right)=\frac{\mathbb E\left[e^{s\xi_{n+1}}\right]}{\mathbb E\left[e^{s\xi_{n+2}}\right]} $$.