Here's how I did it.
Prove that if m | n, then phi(m) | phi(n)
Note that from phi(p^a) = (p^(a-1))(p-1) for some prime p, it follows that phi(p^a) | phi(p^b)
for a | b.
This generalizes to phi(k^a) | phi(k^b) for composite k, since phi is multiplicative.
Since m | n, there exists some q...
There is a crucial problem with your tactic, namely that the Euler phi function is ONLY multiplicative for co-primes.
Consider that 2|8. 8=2*4, but 2 and 4 are not co-prime. Phi(2)=1 and Phi(4)=2. Phi(2)*Phi(4)=2. But Phi(8)=4!
Multiplicativity of Phi alone is sufficient to prove the property...