For every positive integer $m$ there exists a Fibonacci number such that last $m$ digits are $0$

Question

I proved the claim that for every $m\in\mathbb{Z^+}$ the Fibonacci sequence repeats itself for modulo $m$. But how can I move on and make connections with it to prove that for every $m\in\mathbb{Z^+}$ there exists a Fibonacci number such that its last $m$ digits are $0$. Should I show it like using the modulo $10^m$ and show that by the Dirichlet's Box principle after $10^{2m}+1$ it should repeat itself? But does this have anything with my problem?