How can I show that $\phi(m) \mid \phi(n)$? I want to prove that:
$$\text{ if } m,n \geq  1 \text{ and } m \mid n,\text{ then } \phi(m) \mid \phi(n).$$
How can I show this?
I thought the following:
$$m \mid n \Rightarrow \exists k \in \mathbb{Z} \text{ such that } n=km$$
$$\text{We write  } m \text{ like that : }m=p_1^{a_1} \cdots p_k^{a_k}, \text{ where } p_i \text{ are primes and } a_i>0$$
Then,we have $n=k p_1^{a_1} \cdots p_k^{a_k}$
We could show that $\phi(m) \mid \phi(n)$,using the fact that as $\phi$ is multiplicative and if  $(m,n)=1$ ,then $\phi(mn)=\phi(m) \phi(n)$
But..it is possible that $k \mid p_1^{a_1 \cdots p_k^{a_k}}$.
So,do I have to show maybe that $\gcd(kp_1^{a_1} \cdots p_{k-1}^{a_{k-1}},p_k^{a_k})=1$ ?

If the highest power of prime $p_i$ in $m,n$ are $M_i,N_i$ respectively, $0\le M_i\le N_i$  for all $i$
Now for each $M_i>0,$ $$\displaystyle\phi(m)=\prod\phi(p_i^{M_i})=\prod p^{M_i-1}(p_i-1)$$
and similarly for each $N_i>0,$ $$\displaystyle\phi(n)=\prod\phi(p_i^{N_i})=\prod p^{N_i-1}(p_i-1)$$
$\displaystyle p^{M_i-1}$ clearly divides $p^{N_i-1}$ for all $i$ as $M_i\le N_i$  for all $i$  
So, the proposition follows 
