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.

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...

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

Why would the universe care about dimensional analysis? Besides, the outside constant would do the unit conversion from whatever the right-hand side produces to units of force.