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

Proof
Proof by induction:

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

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

Basis for the Induction
$\map P 2$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


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

Then we need to show:


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

Induction Step
This is our induction step:

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

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