If the gcd(s, t)=1, prove $F_s F_t$ divides $F_{st}$ for all s,t > 1
$F_n$ is the Fibonacci sequence, such that $$F_n = F_{n-1} + F_{n-2}$$
I know that it needs to be shown that $F_s F_t\mid F_{st}$ which has the implication that $F_{st} = F_s F_tk$.
I considered that it may be useful to use the theorem that states: the gcd of two fibonacci numbers is also a Fibonacci number with the form $\gcd(F_s,F_t)= g$ where $g= \gcd(s,t)$ or a Corollary relating to this theorem. But I cannot see how this directly applies to the problem.
Is it valid to say that because a theorem states for any $s>1, t>1$, $F_{st}$ is divisible by $F_s$ then that is an implication for $F_s F_t\mid F_{st}$ because the $\gcd(s,t) =1$?