Modular Arithmetic: Additional Lemma

Lemma 1: if a ≡ b (mod p), then ap^n-1 ≡ bp^n-1 (mod pⁿ)

Proof:

(1) a ≡ b (mod p)

(2) So, there exists c such that pc = a - b

(3) So that a = pc + b

(4) So, ap^n-1 ≡ (pc+b)p^n-1 ≡ pⁿc + bp^n-1 ≡ bp^n-1 (mod pⁿ)

QED