Prove that for every NFA N, there is an NFA N' with no epsilon transitions..

In this problem you will prove that e-transitions are merely "syntactic sugar" that makes writing NFAs easier, but e-transitions are not necessary, not even to help reduce the number of states needed. Prove that for every NFA no e-transitions such that (Q... s. F), there is an NFA N'-Q.. .s.P) with L(N'). Note that N' should have the same state set as L(N )