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.