This means that if $\gcd(m, n) = 1$, then $φ(mn) = φ(m) φ(n)$. (Outline of proof: let $A, B, C$ be the sets of nonnegative integers, which are, respectively, coprime to and less than $m, n$, and $mn$; then there is a bijection between $A × B$ and $C$, by the Chinese remainder theorem.)
I also saw this on wiki page of Euler's totient function, but I didn't get the idea$\dots$
My attempts:
By FTA we have: $$a=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$$ $$b=q_1^{\beta_1}\cdots q_m^{\beta_m}$$ Since $\gcd(a,b)=1$, have $p_i\neq q_j$, where $i\in[1,n],j\in[1,m]$, implies: $$a\cdot b=p_1^{\alpha_1}\cdots p_n^{\alpha_n}\cdot q_1^{\beta_1}\cdots q_m^{\beta_m}$$ From this: \begin{align} &~~~~~~\varphi(a\cdot b)\\ &=\varphi (p_1^{\alpha_1}\cdots p_n^{\alpha_n}\cdot q_1^{\beta_1}\cdots q_m^{\beta_m})\tag*{(1)}\\ &=a\cdot b(1-\frac{1}{p_1})\cdots (1-\frac{1}{p_n})(1-\frac{1}{q_1})\cdots (1-\frac{1}{q_m})\tag*{(2)}\\ &=a(1-\frac{1}{p_1})\cdots (1-\frac{1}{p_n})\cdot b (1-\frac{1}{q_1})\cdots (1-\frac{1}{q_m})\tag*{(3)}\\ &=\varphi (p_1^{\alpha_1}\cdots p_n^{\alpha_n})\cdot\varphi(q_1^{\beta_1}\cdots q_m^{\beta_m})\tag*{(4)}\\ &=\varphi(a)\cdot\varphi(b)\tag*{(5)} \end{align}
Is this proof valid, since I saw the proof of Euler's product formula$($used in step $(2))$ seems like also using this property, then if I use Euler's product formula to prove this property, it seems like a little circular, or is there other approaches $?$
