cases even nowadays where "amateurs" find good solutions.

Good point. Even Numberphile and related YouTube channels have highlighted the answer-hidden-in-the-comments on superpermutations (4)(3)(2)(1). And as @standupmaths likes recommending, give it a go, and make a "Parker Square" (5)(6) of it.

Since the number of 11 step solutions becomes infinite by just adding more 1s it's probably not that unlikely to find a solution with several hundreds or thousands digits.*

*: given enough computing power. :-) I mean, adding one 1 will multiply the number of permutations by 16, and from there it will keep on increasing by factors of (15+i)/i . (edit: remove stray factorial, which belied my statement)

You'd need to calculate the density of possible products of single digits numbers in a number range.(easily done with the sieve approach)

...

Otherwise you'd need to prove why it's impossible. ( Which could be done by showing that the density becomes 0)

Hmm. I'll have to think about this some more. I can imagine how I'd start generating the potentials -- 2**$i * 3**$j * 7**$k -- but without generating all the potentials below a threshold, I cannot see any way to generate just potentials of say 200-300 digits. (for any, using the iterators from Higher Order Perl -- section 6.4 especially -- would be how I would start... and brings this reply back into being remotely perl-related :-) ).

Since the number of 11 step solutions becomes infinite by just adding more 1s

I'm also wondering whether there are other base n=11 solutions that aren't permutations of 277777788888899 and however many 1s you insert. A003001 only lists 277777788888899 (since that's the only minimal one), and the other sites I've seen only list that and some of its permutations -- unless I've just not noticed one that looks similar enough to that one that I didn't notice it was separate. OEIS has Numbers with multiplicative persistence value lists for 1-9, but doesn't list the persistence 10 or 11 numbers. But A003001 has a table that (if I understand it correctly) implies there's a second family of persistence=11.

edit 2: found it. A046150 shows 99999999998777772, which would be the family 27777789999999999.