Use induction on 5m

There is a more general result which holds for the Fibonacci Numbers (and some similar sequences): Using $u_{n+1} \equiv u_1u_{n+1} \mod u_n$ and $u_{n+2} \equiv u_{n+1} \equiv u_2u_{n+1} \mod u_n$, we obtain inductively $u_{n+k} \equiv u_ku_{n+1} \mod u_n$. Hence in particular $u_{2n} \equiv u_nu_{n+1} \equiv 0 \mod u_n$ and $u_{3n} \equiv u_{2n}u_{n+1} \equiv 0 \mod u_n$ etc. Inductively, you obtain $\boxed{u_n \mid u_{nm}}$ for all $n,m$ and in particular the case $n=5$ (combined with $u_5=5$) yields the result.