If every Prime-Divisor of m is a prime-Divisor of n ist, then φ(mn) = mφ(n)

Question

To show:

If every Prime-Divisor of m is a prime-Divisor of n ist, then φ(mn) = mφ(n).

φ(n) is the totient-function introduced by euler.

Answer 1

Here is my own answer:

Let $n = p^{a_1}_1 p^{a_2}_2...p^{a_i}_i p^{a_i+1}_{i+1}...p^{a_s}_s$

Every Prime-Divisor of m is a Prime-Divisor of n. So $m = p^{b_1}_1 p^{b_2}_2...p^{b_i}$

n can have more prime divisors than m, thats why it can have the extra $p^{a_i+1}_{i+1}...p^{a_s}_s$.

We know φ(k) = k $\prod_{p|k} (1-1/p)$

so φ(mn) = mn $\prod_{p|mn} (1-1/p)$ $=mn(1-1/p_1)(1-1/p_2)...(1-1/p_i)...(1-1/p_s)$

also φ(n) $= n(1-1/p_1)(1-1/p_2)...(1-1/p_i)...(1-1/p_s)$

so it follows φ(mn) = m φ(n)

q.e.d

Correct?