Here $m,n\in\mathbb{N}\backslash \{0\}$ and $\phi$ is the Euler Totient function. I have been trying to prove that this statement is true but I'm stuck. I started with applying the product formula for $\phi$ to get that the fraction becomes $\prod\_{p\mid n} (1-\frac{1}{p})=\prod\_{q\mid m} (1-\frac{1}{q})$, where $p$ and $q$ are primes. I then did some cross multiplication to obtain that: 
$$\left(\prod\_{q\mid m} (q-1)\right)\cdot\left(\prod\_{p\mid n} p\right)=\left(\prod\_{p\mid n} (p-1)\right)\cdot\left(\prod\_{q\mid m} q\right)$$
Now we know that the largest of the $p\_{i}=p\_{l}$ terms do not divide any of the $p-1$ terms so therefore there is a $q\_{k}=q\_{r}$ that it divides instead, and since both numbers are prime $p\_{l}=q\_{r}$. If we continue this process we find that each $p\_{i}$ is equal to some $q\_{k}$. Now over here is where I got stuck because even though each $p\_{i}$ is equal to some $q\_{k}$ there may be more $q\_{k}$ left over and, so $m$ and $n$ may not have all of the same prime factors. I feel I may be missing something or have done something incorrectly, so any advice would be helpful (thank you in advance).