Good idea, you could do a monte carlo method where you throw many packing peanuts at the cat (or many cats!) and count how many stick, divide by how many you threw, et voila

The universe cares about dimensional analysis because it is invariant under changes of units-of-measure, which are human constructs.

you have free constant k in front of the equation

any dimensional analysis gets consumed by its unknown dimensionality

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q^2n = (-q)^2n which know is ruled out by experiment.

doesn't mean equation can't be using absolute values ("number of electrons/protons") and just applying needed sign at the end

I thought it was "obvious" based on the principle that two charges at the same location should have the same force as one combined charge at that location. Of course this immediately brings up the question of the self-force of a point charge...

Show by experiment that the force of charge0 against charge1+charge2 equals the force of charge0 against charge1 plus charge0 against charge2. Induce an additive-homomorphic property F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2). Then, exponent 1 follows.

I figured this out by listing a bunch of mathematical properties. I couldn't see how the author jumps from zero-preserving to multiply-charges, and I still don't know how, but we can call it out of scope lol

  r : distance between p and q
  q0 : charge 0
  q1 : charge 1
  F : coulomb force function
  charge-commutative: F(r,q0,q1) = F(r,q1,q0)
  zero-preserving: 0 = F(r,q0,0)
  additive-homomorphic: F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2)
  
  homogenous-degree-1: F(r,q0,n*q1) = n*F(r,q0,q1)
  multiplicative-separability: F(r,q0,q1) = K*R(r)*Q(q0,q1)
  multiply-charges: F(r,q0,q1) = K*R(r)*(q0*q1)^a

  Given F(r,q0,q1) = K*R(r)*(q0*q1)^a, charge-commutative, zero-preserving, additive-homomorphic.
  Induction using additive-homomorphic proves homogenous-degree-1. (For example, F(r,q0,2*q1) = F(r,q0,q1+q1) = 2*F(r,q0,q1))
  Equational proof follows from homogenous-degree-1:
  K*R(r)*(q0*n*q1)^a = n*K*R(r)*(q0*q1)^a
  (q0*n*q1)^a = n*(q0*q1)^a
  n^a*(q0*q1)^a = n*(q0*q1)^a
  n^a = n
  n = 0 or a = 1
  n≠0, therefore a=1.

How do you make 3 equal charges by splitting a charge in such a manner?

It's probably easier to verify Gauss's Law, which then gives you Coulomb's law for free.

Direct verification of inverse square laws is hard! For electromagnetic interactions (ie Coulomb's law) you can use scattering. If you want to do a static experiment (Coulomb's Law the hard way or gravity) you probably need a torsion pendulum experiment. AFAIK the best in the world at that are at UW in the Eöt-Wash group: https://www.npl.washington.edu/eotwash/torsion-balances

Could you use an electromagnet and a metal ball to create a fixed charge? If you change the current in the electromagnet, that would change the intensity of the magnetic field. With the metal ball fixed in place nearby and a careful curve for the change of current, you could potentially have a constant charge remaining on the ball.

For two different indivisible charges, this is a bit slippery but I think a=1 by definition. How do we measure charge, practically speaking? By how much force is measured between it and a reference charge. So we take a=1 as a convention. But no experiment can disprove a=2 for indivisible charges as we would simply obtain charge through new units. For assemblies of charges the forces must add linearly due to conservation of momentum, so there we know a=1.