@JonathanAllan hm, I also have another concern with your jelly answer here, specifically that you appear to assume that the cartesian power list is already sorted in a particular order which will return the shortest possible result by taking its head and removing 1s, while it's perfectly possible for [1,2,3,2] to be before [1,4,1,1] for 24!

and so your result to be [2,3,2] instead of [4]

(although for some reason it still returns [4] for that particular example)

hm, one may argue that if [1,4,1,1] appears then [1,1,1,4] will also appear before [1,2,3,2]

so then we can consider [1,1,1,1], [1,1,1,2], [1,1,1,3], ..., [1,1,1,24], [1,1,2,2], ...

while ignoring [1,1,2,1], [1,1,3,1], [1,1,3,2], [1,1,4,1], ..., [1,1,24,23], [1,2,1,1], ..., [1,2,3,2], ...

I think I may use that sneaky trick next time :P

keeping this as proof it works on that part