[hgsgm]

Focusing on v_x(p), the multiplicity of some prime p in the prime factorization of x:

  m = j p^a → v_m(p) = a

  n = k p^b → v_n(p) = b

p is not a factor of j or k

  m^n = j^n (p^a)^n
      = j^n p^(an)
      = k^m p^(bm)
      = …
      = n^m


  v_{m^n}(p) = an = bm

Note that a=0 iff b=0. This is the case where p is neither a factor of m nor n.

  b/a = m/n >=1

This ratio is the same for all primes, so every prime has:

  v_m(p) >= v_n(p)

so

  n | m

(We didn't need the ratio to be the same, but we did need the ratios to be all greater >= 1)