Question 4648
To show :(1 + p)^(p^(n - 1)) = 1 mod p^n ...(**)for all n
 You are right, it relates to binomial theorem (but not Chinese Remainder Thm)
  Proof: Use Math. Induction and Binomial .
   Basic: When n = 1,  (1 + p)^(p^(1 - 1)) = (1 + p)^1 = 1+p mod p^1 = 1.
  Hence, (**) is true when n=1. 
  Induction Hypothesis: When n = k, (1 + p)^(p^(k - 1)) = 1 mod p^k,
   hence, (1 + p)^(p^(k - 1)) = m * p^k + 1.
   Consider  (1 + p)^(p^k) =  (1 + p)^(p^(k-1)*p) = [(1 + p)^(p^(k-1))]^p 
   = (m p^k + 1)^p = E C(p,p-i) (m p^k)^(p-i) 
  [E means summation of i from i =0 to p]
   = m^p p^(k+1) + 1 + E C(p,p-i) (m^(p-i)^ (p^(k*(p-i)) [i runs from 1 to p-1]
   Since p^(k*(p-i)) >= p^(k+1) for all 1<=i<=p-2
   and when i = p-1, p = C(p,p-1) and so p^(k+1) is a divisor of C(p,1)*(m p^k) 
   Hence, (1 + p)^(p^k) = = m^p p^(k+1) + 1 = 1 mod p^(k+1).
   This shows (**) is true when n=k+1. 
   And, the induction proof is complete. 

 Plz read carefully about the details.

 Kenny