How do you rewrite 9^-2 without an exponent

The total amount of trains remaining is equal to the total amount of starting trains that are slower than every train before them (including the first train)

Step-by-step explanation:

Lets do this exercise directly with n trains.

The first train will 'fuse' with the train immediately behind him only if that train is faster. If it is slower, it wont never catch up because the first train will mantain its speed and the second train can only mantain it by fusing itself with a faster train behind it (the speed of the fused train will be the speed of the slower of the 2 trains fused), thus it wont ever adquire the needed speed to catch the first train.

Now, if the first train is slower than the second train, the second train cant lose speed by fusing with trains below him (same argument as above), but since it is faster than the first train, it will eventually fuse with it and the speed will be reduced to the speed of the first train, giving the possibility for other trains to fuse with them too. In short, all consecutive trains starting from the second one and faster than the first train will fuse with the first train.

 The first train slower than the first train wont fuse with this 'chain' of trains, and it will start its own 'chain' of trains, the same way the first train did (it will fuse with any consecutive train faster that it starting from its following train).

This shows that each train slower than the previous trains will form its own chain of trains and it will stay forever. Thus, in the end, the total amount of trains remaining is equal to the total amount of starting trains that are slower than every train before them (including the first train).