Show that $X_0=Y_0$ and that $X_n$ is a martingale.

Suppose $Y_0$ is uniformly distributed on $(0,1]$, and given $Y_n$, suppose $Y_{n+1}$ is uniformly distributed on $(1-Y_n,1]$. Show $X_0=Y_0$, and $$X_n=2^n\prod^n_{k=1}\left[\frac{1-Y_k}{Y_{k-1}}\right],\quad n=1,2,...$$ is a martingale.

I know I need to show that $E[X_{n+1}|\mathcal{F}_n]=X_n$ or $E[X_{n+1}-X_n|\mathcal{F}_n]=0$, whichever one is easier.

$$X_{n+1}=2^{n+1}\prod^{n+1}_{k=1}\left[\frac{1-Y_k}{Y_{k-1}}\right]=2\cdot\left[\frac{1-Y_{n+1}}{Y_n}\right]2^n\prod^n_{k=1}\left[\frac{1-Y_k}{Y_{k-1}}\right]=2\cdot\left[\frac{1-Y_{n+1}}{Y_n}\right]X_n$$ Then since $X_n$ is $\mathcal{F}_n$ measurable, $$E[X_{n+1}|\mathcal{F}_n]=X_nE\left[2\cdot\left[\frac{1-Y_{n+1}}{Y_n}\right]|\mathcal{F}_n\right]$$ I am not sure where to go from here though. I also do not know how to show that $X_0=Y_0$ or why that would be helpful for showing that $X_n$ is a martingale. Could I get some pointers? Thank you.

$E[2\frac {1-Y_{n+1}} {Y_n}|\mathcal F_n]=\frac 2 {Y_n} \int _{1-Y_n}^{1} (1-t)dt/(1-(1-Y_n))=1$. (If you prefer, you can use the change of variable $s=1-t$ to evaluate the integral).