For all $n \geq 1$, and positive integers $a,b$ show: If $\gcd (a,b)=1$, then $\gcd(a^n,b^n)=1$ For all $n \geq 1$, and positive integers $a,b$ show: 
If $\gcd (a,b)=1$, then $\gcd(a^n,b^n)=1$
So, I wrote the $gcd (a,b)=1$ as a linear combination: $ax+by=1$
And, I wrote the $gcd(a^n,b^n)=1$ as a linear combination: $a^n (u)+b^n (v)=1$
can I write the second linear combinations with $x,y$ and then raise the first equation to the nth power or not? 

$$
(a,b)=1\implies ax+by=1
$$
Therefore
$$
\begin{align}
a^nx^n
&=(1-by)^n\\
&=1+b\left(\sum_{k=1}^n\binom{n}{k}b^{k-1}(-y)^k\right)
\end{align}
$$
and
$$
a^nx^n-b\left(\sum_{k=1}^n\binom{n}{k}b^{k-1}(-y)^k\right)=1
$$
Thus, $(a^n,b)=1$. Apply the process again to get $(a^n,b^n)=1$

Another Approach is presented in this answer.
