The congruence equation a≡b(modn)⟹∃k∈Z,a−b=kn.
Taking the i-th power of both sides : (a−b)i=kini.
(a−b)(a−b)i−1=kini.
n∣(a−b),(a−b)∣(a−b)i⟹n∣(a−b)i⟹∃m∈Z,(a−b)i=mn.
But, my proof is incomplete, as it does not show still that ai−bi=kn.
All the comments and answer have ignored the fact that how to get the term ai−bi in the first place.
It seems that all have taken the approach to simply take the ai−bi=kn as a new equality, not derived from the original one. Just it uses the property of n∣(a−b) of the original one.
Answer
It might be easier to use a=kn+b and the Binomial theorem. Then ai=(kn+b)i=∑ij=0(ij)(kn)jbi−j. All of these terms have a factor of n in them except for the 0th one, so this equation may be written in the form ai=pn+bi. Hence the claim.
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