prove that $\operatorname{lcm}(n,m) = nm/\gcd(n,m)$

Hint $\,\ n,m\mid j \!\iff\! nm\mid nj,mj\!$ $\overset{\ \rm\color{darkorange}U}\iff\! nm\mid (nj,mj) \overset{\ \rm \color{#0a0}D_{\phantom |}}= (n,m)j\!$ $\iff\! nm/(n,m)\mid j$

where above we have applied $\,\rm \color{darkorange}U = $ GCD Universal Property and $\,\rm\color{#0a0} D =$ GCD Distributive Law.

Remark $\, $ If we bring to the fore implicit cofactor reflection symmetry we obtain a simpler proof: $ $ it is easy to show $\,d\,\mapsto\, mn/d\,$ bijects common divisors of $\,m,n\,$ with common multiples $\le mn.$ Being order-$\rm\color{#c00}{reversing}$, it maps the $\rm\color{#c00}{Greatest}$ common divisor to the $\rm\color{#c00}{Least}$ common multiple, i.e. $\,{\rm\color{#c00}{G}CD}(m,n)\,\mapsto\, mn/{\rm GCD}(m,n) = {\rm \color{#c00}{L }CM}(m,n).\,$

See here more on this involution (reflection) symmetry at the heart of gcd, lcm duality.

Hint: For any $a,b$ real numbers: $\min(a,b)+\max(a,b)=a+b$.

Now, if we have $a=a_1^{p_1} a_2^{p_2}\ldots$ and similarly with $b$, if you use the equation I just mentioned for all $p_i$, you will get, that $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$.