Show that lcm$(a,b)= ab$ if and only if gcd$(a,b)=1$ Not sure how to begin. If gcd$(a,b)=1$ what can I deduce from that?

Outside of the definition of $gcd(a,b)$ and $lcm(a,b)$, this proof only requires you to know that every integer can be expressed as a unique product of primes. I use a contrapositive proof to prove 
"$\implies$" and a direct proof to prove                     "$\impliedby$"
By contrapositive, assume $gcd(a,b)=d > 1$. Then $a=dr$ and $b=ds$. So $lcm(a,b) \leq drs < (dr)(ds) = ab$
Now assume $gcd(a,b) = 1$. Writing $a$ and $b$ as a product of primes we have:
$a=p_1p_2...p_n$
$b=q_1q_2...q_m$
Where $p$,$q$ are prime and $p_i \neq q_j$ $\forall i,j$ (or else a and b would have a common divisor greater than 1).
Then, any number divisible by a and b must be of the form:
$k(p_1...p_n)(q_1...q_n)$ for some natural number $k$. Then, the smallest such number that is still divisble by a and b is when k=1. So, $lcm(a,b)=ab$.
