Of course, you may calculate higher orders (4 reflects, then travels parallel to 3 - call it 3A, then 3A reflects from bottom surface, making 4A ray, which then refracts to 5A).
Two issues:
1. every such reflection is much dimmer, than previous ones. So the interference pattern from the first order is the most visible;
2. phase shift 5-5A is exactly the same, as as 2-5. Constructive interference occurs when 2-5 phase shift is n*2π. For 5A you'll have 2*n*2π - still positive interference. But - to be honest - this argument works only for positive intereference! For the angles exhibiting negative interference, every second reflection acts opposite.