Proof of a formula involving Euler's totient function: $\varphi (mn) = \varphi (m) \varphi (n) \cdot \frac{d}{\varphi (d)}$ The third formula on the wikipedia page for the Totient function states that $$\varphi (mn) = \varphi (m) \varphi (n) \cdot \dfrac{d}{\varphi (d)} $$
where $d = \gcd(m,n)$.
How is this claim justified?
Would we have to use the Chinese Remainder Theorem, as they suggest for proving that $\varphi$ is multiplicative?
 A: Hint $\ $ A multiplicative function $\rm\:f(n)\:$ satisfies said identity if for all primes $\rm\:p\:$
$$\ \ \ \ \rm\ j\le k\ \Rightarrow\ \ f(p^{j+k}) = \frac{f(p^j)\: f(p^k)\: p^j}{f(p^j)}\ =\ p^j f(p^k)$$
Indeed we have $\rm\ \ \phi(p^{j+k})\ =\ p^{j+k}-p^{j+k-1}\ =\ p^j (p^k-p^{k-1})\ =\ p^j \phi(p^k)$
A: Just as an additional note to the identity provided by @Cardboard Box.
$$
\prod_{p|mn} (1 - \dfrac{1}{p}) = \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})}
$$
Why does this work?
Consider the prime factorization of $m,n$, and that we are multiplying all $p$ such that $p|mn$. But, on first sight, this is the same as multiplying all $p$ such that $p|n$ together with all $p$ such that $p|m$. But what if $m$ and $n$ share a prime factor?
This means that we will multiply that prime factor (call it $p_c$) twice, although of course it can only appear once in the prime factorization of $mn$, albeit with a greater power associated. 
Thus, we need to account for this shared factor, or even shared factors in case there is more than 1 shared prime factor. 
To do this, divide the whole thing by prime factors of the $gcd(m,n)$. Thus, we will be "removing" the factors which we multiplied twice. 
Furthermore, to go from 
$$
mn \frac{\prod_{p|m} (1 - \dfrac{1}{p}) \prod_{p|n} (1 - \dfrac{1}{p})}{\prod_{p|d} (1 - \dfrac{1}{p})} = \phi(m) \phi(n) \dfrac{d}{\phi(d)}
$$
consider multiplying the left side by $\dfrac{d}{d}$. Thus you can group together the $d$ with $\prod_{p|d} (1 - \dfrac{1}{p})$ in the denominator to get $\phi(d)$.
Hope this makes it easier to follow the top answer.
A: You can write $\varphi(n)$ as a product $\varphi(n) = n \prod\limits_{p \mid n} \left( 1 - \frac 1p \right)$ over primes.
Using this identity, we have
$$
  \varphi(mn) 
  = mn \prod_{p \mid mn} \left( 1 - \frac 1p \right)
  = mn \frac{\prod_{p \mid m} \left( 1 - \frac 1p \right) \prod_{p \mid n} \left( 1 - \frac 1p \right)}{\prod_{p \mid d} \left( 1 - \frac 1p \right)}
  = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}
$$
