An adaptation of this answer is given here.

**PDF of a Function of a Random Variable**

If $P(X\le x)=F(x)$ is the CDF of $X$ and $P(Y\le y)=G(y)$ is the CDF of $Y$ where $Y=f(X)$, then
$$
F(x)=P(X\le x)=P(Y\le f(x))=G(f(x))\tag1
$$
Taking the derivative of $(1)$, we get
$$
F'(x)=G'(f(x))\,f'(x)\tag2
$$
where $F'$ is the PDF of $X$ and $G'$ is the PDF of $Y$.

**PDF of the Product of Independent Uniform Random Variables**

If $[0\le x\le1]$ is the PDF for $X$ and $Y=\log(X)$, then by $(2)$ the PDF of $Y$ is $e^y[y\le0]$. The PDF for the sum of $n$ samples of $Y$ is the $n$-fold convolution of $e^y[y\le0]$ with itself. The Fourier Transform of this $n$-fold convolution is the $n^\text{th}$ power of the Fourier Transform of $e^y[y\le0]$, which is
$$
\int_{-\infty}^0 e^{-2\pi iyt}e^y\,\mathrm{d}y=\frac1{1-2\pi it}\tag3
$$
Thus, the PDF for the sum of $n$ samples of $Y$ is
$$
\begin{align}
\sigma_n(y)
&=\int_{-\infty}^\infty\frac{e^{2\pi iyt}}{(1-2\pi it)^n}\,\mathrm{d}t\tag{4a}\\
&=\frac{e^y}{2\pi i}\int_{1-i\infty}^{1+i\infty}\frac{e^{-yz}}{z^n}\,\mathrm{d}z\tag{4b}\\
&=e^y\frac{(-y)^{n-1}}{(n-1)!}\,[y\le0]\tag{4c}
\end{align}
$$
Explanation:

$\text{(4a)}$: take the inverse Fourier Transform

$\text{(4b)}$: substitute $t=\frac{1-z}{2\pi i}$

$\text{(4c)}$: if $y\gt0$, close the contour on the right half-plane, missing the singularity at $z=0$

$\phantom{\text{(4c):}}$ if $y\le0$, close the contour on the left half-plane, enclosing the singularity at $z=0$

We can get the PDF for the product of $n$ samples of $X$ by applying $(2)$ to $(4)$
$$
\bbox[5px,border:2px solid #C0A000]{\pi_n(x)=\frac{(-\log(x))^{n-1}}{(n-1)!}\,[0\le x\le1]}\tag5
$$