Product identity connecting gcd and lcm

Step 1. Write the prime factorizations of $a$ and $b$ over all primes $p$: $a=\prod_p p^{e_p}$ and $b=\prod_p p^{f_p}$, where $e_p,f_p\ge0$ and only finitely many exponents are nonzero.

Step 2. A positive integer $d=\prod_p p^{c_p}$ divides both $a$ and $b$ iff $c_p\le e_p$ and $c_p\le f_p$ for every $p$, so the largest such divisor chooses $c_p=\min(e_p,f_p)$: $\gcd(a,b)=\prod_p p^{\min(e_p,f_p)}$. Dual reasoning for common multiples chooses the smallest exponent at least as large as both, giving $\operatorname{lcm}(a,b)=\prod_p p^{\max(e_p,f_p)}$.

Step 3. For any two real numbers, $\min(e_p,f_p)+\max(e_p,f_p)=e_p+f_p$. Multiplying the two productsprime-by-prime therefore gives $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=\prod_p p^{\min(e_p,f_p)+\max(e_p,f_p)}=\prod_p p^{e_p+f_p}=\left(\prod_p p^{e_p}\right)\left(\prod_p p^{f_p}\right)=ab$.