Product of GCD and LCM

Theorem
$$\mathrm{lcm} \left\{{a, b}\right\} \times \gcd \left\{{a, b}\right\} = \left|{a b}\right|$$

Proof
It is sufficient to prove that $$\mathrm{lcm} \left\{{a, b}\right\} \times \gcd \left\{{a, b}\right\} = a b$$, where $$a, b \in \mathbb{Z}^*_+$$.

Now we have $$a \backslash m \land b \backslash m \Longrightarrow m = a r = b s$$.

Also, $$d = a x + b y$$.

So:

So $$m = n \left({s x + r y}\right)$$.

Thus $$n \backslash m \Longrightarrow n \le \left|{m}\right|$$, while $$a b = d n = \gcd \left\{{a, b}\right\} \times \mathrm{lcm} \left\{{a, b}\right\}$$ as required.

Looks a bit messy to me.