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

Proof
Proof by induction:

For all $r \in \N_{>0}$, let $P \left({r}\right)$ be the proposition:
 * $\displaystyle p \mathrel \backslash \prod_{i \mathop = 1}^r a_i \implies \exists i \in \left[{1 \,.\,.\, r}\right]: p \mathrel \backslash a_i$.

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

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

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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\displaystyle p \mathrel \backslash \prod_{i \mathop = 1}^k a_i \implies \exists i \in \left[{1 \,.\,.\, k}\right]: p \mathrel \backslash a_i$

Then we need to show:


 * $\displaystyle p \mathrel \backslash \prod_{i \mathop = 1}^{k+1} a_i \implies \exists i \in \left[{1 \,.\,.\, {k+1}}\right]: p \mathrel \backslash a_i$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall r \in \N: p \mathrel \backslash \prod_{i \mathop = 1}^r a_i \implies \exists i \in \left[{1 \,.\,.\, r}\right]: p \mathrel \backslash a_i$