This problem outlines an inductive proof of the following theorem that is basically the same as Fermat's Little Theorem.

Fermat's Little Theorem: If p is a prime, then np n (mod p).

(a) State and prove the base case, n = 1.

(b) Now show that if kp k (mod p), then (k + 1)p (k + 1) (mod p).

Please explain your steps verbally (in sentences) so that we can understand your reasoning. You may want to use the following fact about the expansion of polynomials, namely, (x + 1)p = xp + px + . . . + 1. (The dots represent more terms that are divisible by p.) Here are some examples:

(x + 1)2 = x2 + 2x + 1,

(x + 1)3 = x3 + 3x2 + 3x + 1 = x3 + 3(x2 + x) + 1

(x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1 = x5 + 5(x4 + 2x3 + 2x2 + x) + 1.