Consider a probability space $(\Omega, \mathcal{F}, P)$ and random variables $X_0, X_1, \ldots , X_n$ adapted to the filtration $\{\mathcal{F}_t\}_{t\geq0}$. Assume furthermore that each $X_n$ is independent of $\mathcal{F}_{n-1}$.
I want to deduce that if the $X_n$ have non-zero means, the variables \begin{align} M_n = \prod_{k=0}^n \frac{X_k}{\mathbb{E}[X_k]} \end{align} form a martingale. So, \begin{align} \mathbb{E}[M_{n+1} \mid \mathcal{F}_n ] &= \mathbb{E}\big[ \prod_{k=0}^{n} \frac{X_k}{\mathbb{E}[X_k]} \cdot \frac{X_{n+1}}{\mathbb{E}[X_{n+1}]} \mid \mathcal{F}_n \big] \\ &= M_n \cdot \mathbb{E}\big[ \frac{X_{n+1}}{\mathbb{E}[X_{n+1}]} \mid \mathcal{F}_n \big] \\ &= M_n \cdot \mathbb{E}\big[ \frac{X_{n+1}}{\mathbb{E}[X_{n+1}]} \big]. \end{align} I know that, in general, $\mathbb{E}\big[\mathbb{E}[X \mid \mathcal{A}] \big] = \mathbb{E}[X]$. But why is \begin{align} \mathbb{E}\big[ \frac{X_{n+1}}{\mathbb{E}[X_{n+1}]} \big] = 1? \end{align} Is it possible to split the fraction, since $X_{n+1}$ is independent of $\mathcal{F}_n$ \begin{align} \mathbb{E}\big[ \frac{X_{n+1}}{\mathbb{E}[X_{n+1}]} \mid \mathcal{F}_n \big] =^? \frac{\mathbb{E}[X_{n+1} \mid \mathcal{F}_n]}{\mathbb{E}\big[\mathbb{E}[X_{n+1}] \mid \mathcal{F}_n] \big]} = \frac{\mathbb{E}[X_{n+1} ]}{\mathbb{E}[X_{n+1} ]} =1? \end{align}