The distribution of the product of Beta Variables. I need to calculate or at least have an upper bound of
$$\Pr\{B_t > x\}$$ 
where $B_t$ is the product of several Beta variables, i.e. 
$$B_t = \prod_{i=1}^t B(1, k)$$
I've completely no idea how to handle this.  There are countless combinations of these Beta variables that could full fill the condition, so it's hard to get the density or cumulative functions of $B_t$. 
(I happened to know that for uniform variables $\prod_{i=1}^t - \ln U_i \sim \Gamma(t, 1) $, but that seems to be pure luck and could not be generalized to other distributions.)
 A: Note that $B_t$ is the product of $t$ independent random variables distributed like $B_1$. Since $\mathrm P(B_1\geqslant x)=(1-x)^k$ for every $x$ in $(0,1)$, $\mathrm P(B_t\geqslant x)\leqslant \mathrm P(B_1\geqslant x)^t=(1-x)^{kt}$.
A: Since $X \sim \operatorname{B}(1,k)$ is equal in law to $1-U^{1/k}$, where $U$ is uniform on a unit interval, you are computing
$$
   \mathbb{P}\left( \prod_{i=1}^t (1-U_i^{1/k}) > x \right)
$$
The case of $k=1$ is special, due to $1-U \stackrel{d}{=} U$, and $-\log(\prod_{i=1}^t U_i) \sim \Gamma(t,1)$, as you mention.
Observe that $m_r = \mathbb{E}\left( B_t^r \right) = \mathbb{E}\left( B_1^r \right)^t =  \binom{k+r}{r}^{-t}$. This permits to compute the moment generating function for $B_t$ in terms of the confluent generalized hypergeometric function:
$$
   \mathcal{M}_{B_t}(u) = \mathbb{E}\left( \mathrm{e}^{u B_t} \right) = \sum_{r=0}^\infty \frac{u^r}{r!} m_r = {}_t F_t\left(\left. \begin{array}{c}\underbrace{1,\ldots,1}_{t\text{ times}} \\ \underbrace{k+1,\ldots,k+1}_{t\text{ times}} \end{array}\right| u \right)
$$
So in principle one can construct Chernoff bound. But it is known that moment bounds are tighter. Assuming $0 < x<1$,
$$
   \mathbb{P}(B_t \geqslant x) \leqslant \inf_{r\geqslant 0} \left( \frac{m_r}{x^r}\right) = \inf_{r\geqslant 0} \left( \frac{1}{x^r} \binom{k+r}{r}^{-t}\right) = 
  \left. \frac{1}{x^r} \binom{k+r}{r}^{-t}  \right|_{r = \lfloor k \left(\frac{1}{1-x^{\frac{1}{t}}}-1\right)-1 \rfloor}
$$

