Distribution of the product of two iid variables with cdf $x^n$ on [0,1] I am following the Wikipedia Proof
but something goes wrong:
$$ f_X(x) = a x^{a-1} \text{ for } x \in [0,1]$$
If $X$ and $Y$ are iid, I want the distribution of $XY$. Actually I want it for the product of $n$ such iid variables, but I thought I would derive that by starting with two.  Yet when I try to follow the procedure:
$$f_{XY}(z) = \int_{0}^{1}f_X(x)\cdot f_Y(z/x)\cdot \frac{1}{|x|} dx $$ it diverges.
$$\int_0^1 ax^{a-1} \int_0^{z/x}ay^{a-1} dy dx = \int_0^1 az^a\cdot \frac{1}{x} dx$$
I tried following it step by step but I don't see my mistake.  Yet it seems obvious the distribution should be something reasonable.  Where is my error?
Ultimately I am trying to find $$E\Big[\frac{-n}{\ln(\Pi X_i)}\Big]$$ but I got it down to needing the distribution of $\Pi X_i$.
 A: 
but I got it down to needing the distribution of $\prod_i X_i$.

If your goal is to find the expectation you showed, the distribution of you are trying to derive is not needed.
In fact, observe that the distribution of
$$Y=-\log X$$
is....after easy calculations
$$f_Y(y)=a e^{-ay}$$
a negative exponential (nice result :)).
now let's take you expectation
$$\mathbb{E}\left[ \frac{-n}{\log\prod_i X_i} \right]=\mathbb{E}\left[ \frac{n}{\sum_i(-\log X_i)} \right]=n\mathbb{E}\left[\frac{1}{Z}\right]=\frac{n}{n-1}\alpha$$
this because $(1/Z)\sim \text{Inverse Gamma}$
If you do not want to use the known result of the inverse gamma  you can calculate $\mathbb{E}\left[\frac{1}{Z}\right]$ by integration using the fact that $Z\sim\text{Gamma}(n;a)$
Z in fact is the sum of $n$ iid negative exponentials

Expectation's calculation by integration
$$n\mathbb{E}\left[\frac{1}{Z}\right]=n\int_0^{\infty}\frac{1}{z}\frac{\alpha^n}{\Gamma(n)}z^{n-1}e^{-\alpha z}dz=$$
$$=\frac{n \alpha\Gamma(n-1)}{\Gamma(n)}\underbrace{\int_0^{\infty}\frac{\alpha^{n-1}}{\Gamma(n-1)}z^{(n-1)-1}e^{-\alpha z}dz}_{=1}=\frac{n}{n-1}\alpha$$

Further conclusion (not requested but probably yes...)
your estimator is biased for $\alpha$ but asymptotically unbiased.
