elementary number theory – Show $ gcd (a, b) = 1 $ implies $ varphi (a cdot b) = phi (a) cdot phi (b) $

This means that if $ gcd (m, n) = 1 $, then $ φ (mn) = φ (m) φ (n) $, (Evidence: let $ A, B, C $ are the sets of nonnegative integers that are respectively too and less than coprime $ m, n $, and $ Mn $; then there is a bijection between $ A × B $ and $ C $, according to the Chinese remainder theorem.)

I also saw this on the wiki page of Euler's Totient function, but I had no idea$ dots $

My experiments:

After 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 $, to 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 that:
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

Is this proof valid since I have seen the proof of Euler's product formula?$ ($used on step $ (2)) $ It seems like I would use this feature too. If I then use Euler's product formula to prove this property, it seems a bit circular, or are there other approaches $? $