Let's represent two integer (\(a\) and \(b\)) as products of prime: \(a={p_1}^{a_1}{p_2}^{a_2} \ldots {p_n}^{a_n}\) and \(b={p_1}^{b_1}{p_2}^{b_2} \ldots {p_n}^{b_n}\), where \(p_n\) is the largest prime that divides \(max\{a,b\}\). It's possible that \(a_i\) or \(b_i\) could be 0 (i.e the prime \(p_i\) is not a factor). The gcd can be represented like this:
$$ \gcd (a,b) = {p_1}^{min \{a_1, b_1\} } {p_2}^{min \{a_2, b_2\} } \ldots {p_n}^{min \{a_n, b_n\} }$$
The lowest common multiple of both \(a\) and \(b\) is the smallest positive integer that both \(a\) and \(b\) can divide. It can also be represented as:
$${p_1}^{max \{a_1, b_1\} } {p_2}^{max \{a_2, b_2\} } \ldots {p_n}^{max \{a_n, b_n\} }$$
Now let's multiply both gcd and lcm:
$$\begin{align} (a, b) &* [a, b] \\ {p_1}^{min \{a_1, b_1\} } {p_2}^{min \{a_2, b_2\} } \ldots {p_n}^{min \{a_n, b_n\} } &* {p_1}^{max \{a_1, b_1\} } {p_2}^{max \{a_2, b_2\} } \ldots {p_n}^{max \{a_n, b_n\} } \end{align}$$
We can rewrite this as:
$$ ({p_1}^{min \{a_1, b_1\} }{p_1}^{max \{a_1, b_1\} }) ({p_2}^{min \{a_2, b_2\} }{p_2}^{max \{a_2, b_2\} }) \ldots ({p_n}^{min \{a_n, b_n\} }{p_n}^{max \{a_n, b_n\} }) $$$$ ({p_1}^{min \{a_1, b_1\} max \{a_1, b_1\} }) ({p_2}^{min \{a_2, b_2\} max \{a_2, b_2\} }) \ldots ({p_n}^{min \{a_n, b_n\} max \{a_n, b_n\} }) $$
Since \( min \{ a_i, b_i \} max \{ a_i, b_i \} = a_i b_i \):
$$ {p_1}^{a_1 b_1 } {p_2}^{a_2 b_2 } \ldots {p_n}^{a_n b_n } $$$$ ({p_1}^{a_1 } {p_2}^{a_2 } \ldots {p_n}^{a_n}) ({p_1}^{b_1 } {p_2}^{b_2 } \ldots {p_n}^{b_n }) $$
The above is equal to \(ab\). Therefore, \((a,b)[a,b] = ab\).