Euclid's Lemma for Prime Divisors/General Result/Proof 2

Proof
Proof by induction:

For all $r \in \N_{>0}$, let $P \paren r$ be the proposition:
 * $\displaystyle p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$.

$P(1)$ is true, as this just says $p \divides a_1 \implies p \divides a_1$.

Basis for the Induction
$P(2)$ is the case:
 * $p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$

which is proved in Euclid's Lemma for Prime Divisors.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \paren k$ is true, where $k \ge 1$, then it logically follows that $P \paren {k + 1}$ is true.

So this is our induction hypothesis:


 * $\displaystyle p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$

Then we need to show:


 * $\displaystyle p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$

Induction Step
This is our induction step:

So $P \paren k \implies P \paren {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$