2019/005) Prove If $gcd(a,b)=1$ then $gcd(a,b^2)=1$

We have $ax + by = 1$  as per bezout identity 

So$b = 1 * b = b* (ax + by)$ 
$= bax + b^ 2 y$ 
So $by = baxy + b^2y^2$ 
Or $by + ax = baxy + ax + b^2 y^2$ 
Or $1 = ax(1+by) + b^2 y^2$ 

As we can put 1 as linear combination of $a$ and $b^2$ so $gcd(a,b^2) = 1$