Let \(m = p_1 ^ {b_1} p_2 ^ {b_2} \cdots p_h ^ {b_h}\), where \(p_i\) is a prime factor of \(m\) and \(b_1 \ge 1\), and let \(n = p_1 ^ {a_1} p_2 ^ {a_2} \cdots p_k ^ {a_k}\), where \(p_i\) is a prime factor of \(n\) and \(a_1 \ge b_i\).
\(p_i\) are primes not in any specific order, so it's possible that \(p_1 \gt p_2\) and it's possible that \(p_2 \gt p_1\). The important part to note is that the prime factors of \(m\) (which are \(p_i\) where \(i \le h\)) are also prime factors of \(n\), so \(m|n\).
The formula for \(\phi(n)\) (derived here) is:
All terms in \(\prod_{i=1}^h (1 - 1/p_i) \) also exists in \(\prod_{i=1}^k (1 - 1/p_i)\). So \(\prod_{i=1}^h (1 - 1/p_i) \) divides \(\prod_{i=1}^k (1 - 1/p_i)\). Since \(m\) divides \(n\) as well, then all terms in \(\phi(m)\) divides the terms in \(\phi(n)\). Therefore, \(\phi(m) | \phi(n)\).