CDF of the product of $n$ i.i.d random variables given by $f(x)=\frac1{a+1} x^a$ The random variables $X_1,\ldots,X_n$ are i.i.d and have the distribution $f(x)=(a+1) x^a$ on $[0,1]$. 
I want to find a formula for $$F(t)=\operatorname{Pr}[X_1\cdots X_n \leq t]= {(a+1)^n}\int_{x_1,\ldots,x_n \in [0,1] \;\land \; x_n\leq \frac{t}{x_1\cdots x_{n-1}}} x_1^a\cdots x_n^a \ dx_1\cdots dx_n$$
where $t \in [0,1]$. This integral does not seem to split easily to an iterative integral because of the interleaved dependence between the variables. Any trick here?
 A: Perhaps the easiest way of doing this is to show that $Y = \log X_i$ is an exponential random variable, or rather it is the negative of an exponential random variable. One way to get this is to calculate the density given by
$$
\begin{align*}
f_{Y}(y) &= (a+1)\left( e^y \right)^a \left| \frac{d e^{y}}{d y} \right| \\
&= (a+1) e^{(a+1)y},
\end{align*}
$$
and noting that $y \leq 0$. You can now use the fact that the sum of $n$ exponential random variables with common rate parameter $\lambda$, is a Gamma$(n, \lambda^{-1})$ random variable (using the shape/scale parameterisation). In particular let
$$
S = \sum_{i=1}^{n} \log X_i, \qquad -S \sim \mbox{Gamma}(n,1/(a+1)) 
$$
and so in particular we get
$$
\begin{align*}
\mathbb{P}\left[ \prod_{i=1}^n X_i \leq z \right] &= \mathbb{P}\left[\sum \log X_i \leq \log z \right] \\
&= 1 - \frac{1}{\Gamma(n)}\gamma\left(n,-\frac{\log z}{a + 1}\right),
\end{align*}
$$
where $\gamma$ is the usual lower incomplete Gamma function. 
A: Here's something that works when $n=2.$
\begin{align}
u & = x_1 x_2 \\
v & = x_1/x_2 \\[10pt]
\text{So } x_1 & = \sqrt{uv} \\
\text{and } x_2 & = \sqrt{u/v \, } \\[10pt]
\text{and } dx_1\,dx_2 & = \left| \dfrac{\partial(x_1,x_2) }{\partial(u,v)} \right| = v+1.
\end{align}
$$
\iint\limits_{\text{region}} x_1^a x_2^a \, dx_1 \, dx_2 = \int_0^1 \left( \int_u^{1/u} u^a (v+1) \, dv \right) \, du = \text{etc.}
$$
Next I would try to figure out whether that can be adapted to $n>2.$
