I'm working in the following Euler's theorem exercise:

> 
> if $d|n$ then $\phi(dn)=d\phi(n)$
> 
> 
> 

Hints for a shorter and simpler, imo, approach:

I'm based on:

$$n=\prod\_{i=1}^k p\_i^{a\_i}\;,\;\;p\_i\;\;\text{primes}\;\;,\;\;a\_i\in\Bbb N\implies \phi(n)=n\prod\_{i=1}^k\left(1-\frac1{p\_i}\right)$$

So now:

$$d\mid n\implies n= xd\;,\;\;x\in\Bbb N\implies\; \phi(x)=x\prod\_{i=1}^k\left(1-\frac1{p\_i}\right)$$

I'm not sure this is the right way, how should I involve that $d$ so I would be able to show the requested property? Any help will be really appreciated.