If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$ 
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$

This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..

Lemma: $\gcd(a,b)=1\implies\gcd\left(a^2,b^2\right)=1$
Proof: By Bezout, $\gcd(a,b)=1$ implies
$$
\begin{align}
ax+by&=1\\
a^2x^2&=1-2by+b^2y^2\\
\left(2y-by^2\right)b&=1-a^2x^2\\
\left(2y-by^2\right)^2b^2&=1-2a^2x^2+a^4x^4\\
\left(2x^2-a^2x^4\right)a^2+\left(2y-by^2\right)^2b^2&=1
\end{align}
$$
Therefore, $\gcd\left(a^2,b^2\right)=1$.$\qquad\square$
Corollary: $\gcd(a,b)=1\implies\gcd\left(a^{2^n},b^{2^n}\right)=1$
Proof: Induction using the Lemma.$\qquad\square$
Thus, for any $k$, we can find an $n$ so that $k\le2^n$. The Corollary says that there are $x_n$ and $y_n$ so that
$$
a^{2^n}x_n+b^{2^n}y_n=1
$$
and therefore,
$$
a^k\left(a^{2^n-k}x_n\right)+b^k\left(b^{2^n-k}y_n\right)=1
$$
Thus, $\gcd(a,b)=1\implies\gcd\left(a^k,b^k\right)=1$.
