It is well known that every number $x$ in the unit interval $[0,1]$ is the arithmetic mean of two elements of the (triadic) Cantor set $C$. The way to see it I like the most: the Cantor set is the unique fixed point of a self-map on the set of non-empty closed subsets of $[0,1]$, namely $\kappa: F\mapsto \frac13F+\big\{0,\frac23\big\}$; this map $\kappa$ is a contraction with respect to the Hausdorff distance, which is complete, and the subset of all closed sets $F$ such that $F+F=[0,2]$ is a closed non-empty $\kappa$-invariant set, so it contains the fixed point. (So, incidentally, if $C':=C-\frac12$, we have $C'-C'=[-1,1]$, which gives a counterexample to the converse of the Steinhaus theorem, stating that $A-A$ is a neighborhood of $0$ provided that $A$ has positive Lebesgue measure.)

The analogous result for products of two elements of $C$ is not true: already $C_1:=[0,\frac13]\cup[\frac23,1]$ happens to give a set of products with a gap: $C_1\cdot C_1=[0,\frac13]\cup[\frac49,1]\subsetneq[0,1]$. Things seem to go better for products of three Cantor set's numbers, and I would believe that they actually cover the whole unit interval, since it turns out that $C_1\cdot C_1\cdot C_1=[0,1]$. However, this is not sufficient to conclude the result for $C$: I'm not sure if e.g. the set of all closed sets $F\subset[0,1]$ such that $F+F=[0,2]$ and $F\cdot F\cdot F=[0,1]$ is actually $\kappa$-invariant.

Questions: is every $x\in [0,1]$ a product of (finitely many) Cantor set's elements? If so, is there a uniform bound on the set of factors? And, if so, what is the minimum $n$ such that any $x\in [0,1]$ can be expressed as a product of $n$ numbers in the Cantor set?