How exactly does a transistor amplify the input current in CE configuration
This is grossly simplified, but I think you're asking for a very simple answer, so take it in the spirit it's offered. 

The key lies in the fact that for some semiconductors/junctions,

a) you can talk about majority and minority charge carriers, roughly electrons and holes, depending on whether the material is p-type or n-type.

b) under a broad range of conditions, the product of minority and majority carriers remains constant.

This means that, for the proper materials and junction configurations, such as a bipolar junction transistor (BJT) when you inject a small number of minority carriers, you make a large difference in the total minority population. Let's say you double the minority concentration by adding some current. Since the product of minority and majority carriers remains constant, if you double the number of minority carriers, you must halve the number of majority carriers, which provide current to the device outputs. Since, by definition, minority carriers are in the minority (and by a large proportion in practical devices), a small change in their concentration produces a large change in the concentration of the majority carriers. In other words, the injected current gets amplified.

Like I say, this is the really, really simplified version, and is just one step up from Lies-To-Children. If you need details or a better understanding, you're just going to have to learn to wade through the equations. Sorry.