Question: How do I find the pdf of $\prod\_{i=1}^n X\_i$, where $X\_is$ are independent uniform [0,1] random variables.

I know X~U[0,1], -ln(x) is exponential(1).
I also know the sum of two or more independent exponential random variable is gamma. 

For $Y = \sum\_{i=1}^n -ln(x\_i)$, which is a gamma(n, 1), I found the pdf for Y is $$\int\_0^\infty \frac{1}{\Gamma (n)} y^{n-1} e^{-y} dy$$.

Let $Z = e^Y$

I am trying to the pdf for Z, what I found is $$\int\_1^\infty \frac{1}{\Gamma (n)} ln(z)^{n-1} \frac{1}{z^2} dz$$, which does not look right to me.
Could someone check it?

Answer: 
Step 1: Note that $-\log X\_j$ has an exponential(1) distribution.

Step 2: Note that $\sum\_{j=1}^n (-\log X\_j)$ is a sum of iid random variables, so convolution results apply.

Step 3: Figure out the convolution for the exponential distribution, by looking it up, by using characteristic functions, or moment-generating functions, whichever you're familiar with. You'll find $f\_X(x)=x^{n-1} e^{-x} / (n-1)!$ as the density of $-\log Y$, which is a $\Gamma$ distribution.

Step 4: Transform this back to get something like $(-\log y)^{n-1}/(n-1)!$. 

Note that for increasing $g$ (decreasing $g$ is analogous),
$$\Pr(Y\leq y)=\Pr(g(X)\leq y)=\Pr(X\leq g^{−1}(y))=F\_X(g^{−1}(y)).$$

Now differentiate with respect to $z$ to get 

$$
f\_Y(y) = f\_X(g^{-1}(y)) g^{-1}\;'(y)= \frac{f\_X(g^{-1}(y))}{g'(g^{-1}(y))}.
$$

In your case $g(x)=\exp(−x)$ which is decreasing in $x$ and hence you need to replace the denominator with $|g′(g^{−1}(y))|$.

So $g'(x)=-\exp(-x)$ and $g^{-1}(y)=-\log y$.
An attempt:

For the case $n=2$, using directly the fact that (for $Y=X\_1X\_2$, where $X\_1,X\_2\sim\operatorname{Unif}([0,1])$ are independent)
$$
\forall z\in[0,1],\quad f\_Y(z)=\int\_0^1 f\_{X\_1}(x)f\_{X\_2}\left(\frac{z}{x}\right)\frac{dx}{x}
$$
you get that for all $z\in(0,1]$
$$
f\_Y(z)=\int\_0^1 1\cdot \mathbb{1}\_{\{\frac{z}{x}\in[0,1]\}}\frac{dx}{x} = \int\_z^1 \frac{dx}{x} = \ln\frac{1}{z}
$$

For the general case, you can apply the trick of taking the logarithm of $Y=\prod\_{k=1}^n X\_k$, i.e. consider $Z\stackrel{\rm def}{=} \ln Y = \sum\_{k=1}^n \ln X\_k = \sum\_{k=1}^n Z\_k$ for $Z\_k\stackrel{\rm def}{=} \ln X\_k$, so that for all $z\in(-\infty,0]$
$$
f\_Z(z)=(f\_{X\_1}\ast\cdot\ast f\_{X\_k})(z) = (f\_{X\_1}\ast\cdot\ast f\_{X\_1})(z)
$$
where $\ast$ is the convolution operator and
$$
\begin{align\*}
f\_{X\_1}(x) &= \frac{d}{dx}\int\_{-\infty}^x \mathbb{P}\left\{\ln X\_1 \in dt\right\} = \frac{d}{dx} \mathbb{P}\left\{\ln X\_1 \leq x\right\}= \frac{d}{dx} \mathbb{P}\left\{X\_1 \leq e^x\right\}\\
&= \frac{d}{dx}\mathbb{P}\left\{X\_1 \leq e^x\right\}= \frac{d}{dx} e^x = e^x
\end{align\*}
$$
(caveat: I haven't carefully checked my calculations) Once (after a lot of fun) you figure $f\_Z$, you can go back to $f\_Y$ by a similar process (using the cdf of $f\_Z$, etc).