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? 

You can use the unique prime factorization to show this result logically . If $a$ & $b$ are relatively prime then they have no common prime factors. Therefore any power of $a$ & $b$ will just repeat each of the prime factors $n$ times. Therefore $a^n$ and $b^n$ still have no common prime factors and therefore are relatively prime.  
