Is this proof valid? $P|a^k \implies P|a$

Question

I'm trying to prove $P|a^k \implies P|a$

$p$ is a prime, $a \in \mathbb{Z}^{+} \\$, $n \geq 2 \in \mathbb{Z}$

I know that: $P|ab \implies P|a \text{ or } P|b \dots [1]$

a is a positive integer, b is a positive integer

Base: $P|a^2 \implies P|a$

$P|aa \implies P|a$ using [1]

Induction hypothesis: Assume $P|a^k \implies P|a$

Check: $P|a^{k+1} \implies P|a$

$P|aa^k \implies P|a$ using [1]

therefore $P|a^k \implies P|a$

Answer 1

It is not correct. From $P\mid a^{k+1}=a\times a^k$, what you can deduce is that $P\mid a$ or that $P\mid a^k$. And then you can use the induction hypothesis.