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? 

There are many ways to prove this, depending on one's background and the desired generality. Below are a few, some well-known, some not. First, the most obvious is to use the fundamental theorem of arithmetic, i.e. existence and uniqueness of prime factorizations, e.g. as follows:
Hint $\ $ prime $\,p\:|\:a^n,b^n\:\Rightarrow\:p\:|\:a,b,\:$ since prime $\:p\:|\:d_1\cdots d_k\:\Rightarrow\:p\:|\:d_1\ $ or $\,\ldots\,$ or $\:p\:|\:d_k\,$   
Alternatively, it may be derived as  a simple inductive consequence of Euclid's Lemma, i.e. $\,(a,b)=1=(a,\,c)\ \color{#c00}{\Rightarrow}\,(a,bc)=1.\,$ Hence for $\,c=b\,$ we infer $\,(a,b^2)=1,\,$ and by induction $\,(a,b)=1=(a,b^n)\color{#c00}\Rightarrow\,(a,b^{n+1})=1.\,$ Reversely, inducting in the same way on powers of $\,a,\,$
$\,(b^n,a) = 1\,\color{#c00}\Rightarrow\, (b^n,a^2) = 1\,\ldots\,\color{#c00}\Rightarrow\, (b^n,a^k) = 1.$
Alternatively, Gauss's Lemma (GL) yields a quick proof. Let $\rm\:{\cal C}(f)\:$ denote the content of a polynomial, i.e. the gcd of its coefficients. GL states $\rm\: {\cal C}(f\,g)\ =\ {\cal C}(f)\ {\cal C}(g)\ $ hence
$\rm\qquad\qquad\qquad\ \ 1\ =\ (a,b)\ =\ {\cal C}\:(a\ x + b)\ =\  {\cal C}\:(a\ x - b)$
$\rm\qquad\qquad \Rightarrow\ \ 1\ =\ {\cal C}\:((a\ x + b)\:(a\ x - b))\ =\ {\cal C}\:(a^2\: x^2 - b^2)\: =\: (a^2,b^2)$ 
Iterating shows $\rm\,(a^n,b^n) = 1\,$ for $\rm\,n = 2^k,\,$ hence for all $\rm\:n,\:$ by $\rm\,m\le n\,\Rightarrow\,(a^m,b^m)\:|\:(a^n,b^n),\,$ another example of the "up then down" (or interval) induction.
More generally, one easily proves the Freshman's Dream $\,(a,b)^n = (a,b)^n\,$ for gcds or invertible ideals, by simple high-school arithmetic using their familiar arithmetical laws (associative, commutative, distributive,cancellative) along with the more special idempotent law $(a,a) = (a).$ The Bezout-based proofs in the answer of André and Rob are special cases of this method.
