@ToddTrimble "chaology" looks more like English Greek to me than "chaosology", which looks like a formation within English, taking the suffix to be "-ology". The reason why I say English Greek is that in Greek, words such as analogy are αναλογια, which would be analogia, so the final y is of purely English origin. Chaology comes from "chaos" and "logos" in an analogous way to how topology comes from "topos" and "logos" and genealogy comes from "genea" and "logos". Taking "-ology" as the suffix instead of "-logy" came later.

There is also "puissance" in French, which can often be seen in old articles by Kuratowski or Sierpinski.

The natural cognate of "Mächtigkeit" in English is "mightiness", though it seems to have never been used as a translation, "power" being used instead, as pointed out by David Roberts. An example of an old book where it is used is: books.google.dk/… The phrase "power of the continuum" also seems to have lasted a bit longer than "power" as a general term for cardinality.

@TomWerner Can you show me an explicit example where the projection has more faces? I can't think of one.

@TomWerner Quite right, I've just made the edit.

@DavidHarris A set-theoretic analogue of undefined behaviour is asking whether $2^{\aleph_0} = \aleph_1$. This is anything but junk, however -- it's just that the "standard" (i.e. ZFC axioms) does not define cardinal exponentiation.

@Number Yes they would, at least to me. I was introduced to umbral calculus by Rota's version of it, so I've not seen the classical literature.

You'll need infinitely many of the groups to be nontrivial. The denumerable product of the trivial group is the trivial group. Also, it seems that the group structure isn't playing any role here, it's just the fact that a denumerable product of finite sets of cardinality $\geq 2$ is homeomorphic to the Cantor space by Brouwer's characterization of it as a compact metrizable totally disconnected space with no isolated points.

@AryehKontorovich That was loose wording by me (I have edited the answer). What I mean is the existence of such a cardinal being consistent.

@AryehKontorovich On my side it looks like my counter-edit won, so there is no problem with the answer as it is now.

@AryehKontorovich The $\aleph_1$ is intentional. It's a strict inequality. Otherwise the definition of $\kappa$-additivity doesn't work in the definition of a measurable cardinal (the measure is only additive for disjoint families strictly smaller than $\kappa$). Sets of cardinality $< \aleph_1$ are exactly countable sets.