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

Hint $\,\ n,m\mid k \!\iff\! nm\mid nk,mk\!\overset{\ \rm U}\iff\! nm\mid (nk,mk) \overset{\ \rm D_{\phantom |}}= (n,m)k\!\iff\! nm/(n,m)\mid k$

where above we have applied $\rm U = $ GCD Universal Property and $\rm D =$ GCD Distributive Law.

Remark $\ $ If we bring to the fore the implicit reflection symmetry we obtain a simpler proof: it's easy to show that $\,d\,\mapsto\, mn/d\,$ bijects the common divisors of $\,m,n\,$ with the 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 $\gcd(a,b)\cdot\operatorname{lcm}(a,b)=ab$.