Hint: $\max \lbrace X_1,...,X_N \rbrace \leq x$ if and only if $X_i \leq x$ for all $i=1,...,N$.
Let $\Sigma_1=\sum_{i=1}^{\infty}{(\prod_{j=1}^{j}{X_i})}$ and notice that
$\Sigma_1=X_1(1+\sum_{i=2}^{\infty}{(\prod_{j=2}^{j}{X_i})})=X_1(1+\Sigma_2)$
Also note that $\Sigma_1$ and $\Sigma_2$ are also iid (can be seen via dummy index changes).
Let us also assume that $\Sigma_i$ has the pdf $f_{\Sigma}(\sigma)$
Define $Y=(1+\Sigma_2)$, $f_Y(y)=f_{\Sigma}(y-1)$
Anyway since $\Sigma_2$ does not contain $X_1$, they are also independent.
Finally we can write down:
$f_{\Sigma}(\sigma)=\int_{\mathscr{Y}}{f_X(\sigma/y)f_Y(y)dy}=\int_{x=0}^{1}{f_X(x)f_Y(\sigma/x)dx}=\int_{x=0}^{1}{f_X(x)f_{\Sigma}(\sigma/x-1)dx}$
For an arbitrary $f_X(x)$ this cannot be solved, even existence of $f_{\Sigma}$ cannot be guaranteed.
One trivial solution for $(X,\Sigma)$ pair is satisfied by $f_X(t)=f_\Sigma(t)=\delta(t)$
One trivial $X$ implying that $\Sigma$ does not have a (not heavy tailed) pdf is: $f_X(t)=\delta(t-1)$
So if you can give us $f_X(x)$, we can talk again about the solution. If not for all I care I have given you a solution and an answer.
Best Answer
$X_n$ is a deterministic function of $X_1,\ldots, X_{n-1}$ so you're basically done. You can express it as "strings where the first $n-1$ bits are uniform over $\{0,1\}^{n-1}$ and where the $n$th bit is the number of $1$s in the first $n-1$ bits."
I'm guessing the more "elegant" expression would be "uniform over all $n$-bit strings with an even number of $1$s."