Prove that $[n, n + 1] = n × (n + 1)$ where $n$ is a natural number.
We'll use Proposition (2.22).
Let $a$ and $b$ be positive integers then $\gcd (a, b) × [a, b] = a × b$.
So we have
$$ [n, n+1] = \frac{n \times (n+1)}{\gcd(n, n+1)} $$
We showed in Exercise (2.1).6 that that $\gcd(n, n+1)=1$, and so the desired result follows immediately
$$ [n, n+1] = n \times (n+1)$$