Divisibility of Exponents

Question

So I'm having trouble trying to show this,

a,b and x are positive integers. If $a\mid b^x$, show that some factor $k$ of $a$ divides $b$.

In other words, if a number $a$ divides a power, how can I show that some factor of $a$ (or $a$ itself) divides $b$.

Answer 1

$b^x=b.b..x times$ now as $a|b, ^x$ so there is always a factor of $a$ which will divide $b$ you can assume $a|b^x=m$ so $b^x=ma$ and now continue . I don't think there's much more.

Answer 2

Case $a=1$ is trivial. Let's assume $a>1$, then this is equivalent to prove that $\gcd(a,b) \neq 1$.

If $\gcd(a,b)=1$, it's easy to see that $\gcd(a,b^x)=1$, which is contradictory to $a \mid b^x$. Let $k=\gcd(a,b)$, then $k$ is factor of $a$ that divides $b$.