Show that if $a^n ≡ 0 \pmod p$ where $p$ is prime then $a ≡ 0 \pmod p$.
We will use induction on $n$.
Let $P(n)$ be the statement
$a^n \equiv 0 \pmod p \implies a \equiv 0 \pmod p$ for prime $p$.
We need to prove a base case $P(1)$ and an inductive step $P(n) \implies P(n+1)$.
Base Case
The base case is
$a^1 \equiv 0 \pmod p \implies a \equiv 0 \pmod p$ for prime $p$.
The antecedent and the consequent are equivalent statements so the base case is true.
Inductive Step
We assume the induction hypothesis $P(n)$ and aim to show $P(n+1)$, which is
$a^{n+1} \equiv 0 \pmod p \implies a \equiv 0 \pmod p$ for prime $p$.
Noting that $a^{n+1} = a^n \times a$, we have
$$ a^{n+1} = a^n \times a \equiv 0 \pmod p $$
Proposition 3.14 gives us $a^n \equiv 0 \pmod p$ or $a \equiv 0 \pmod p$. Let's consider both cases in turn.
Case 1: $a^n \equiv 0 \pmod p$, which by the induction hypothesis, implies $a \equiv 0 \pmod p$.
Case 2: $a \equiv 0 \pmod p$ is the conclusion we desire.
In both cases we conclude $a \equiv 0 \pmod p$, and so $P(n+1)$ is true.
We have shown by induction that if $a^n ≡ 0 \pmod p$ where $p$ is prime then $a ≡ 0 \pmod p$.