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..

The proof is by induction. Let us denote $\gcd(a,b) = d$.
For $n=1$, there is nothing to prove. Assume that it is true for all $n \leq k$ i.e. $\gcd(a^n,b^n) = d^n$ for all $n \leq k$. We now need to prove that $\gcd(a^{k+1},b^{k+1}) = d^{k+1}$. Since $\gcd(a^k,b^k) = d^k$, there exists $x_k,y_k \in \mathbb{Z}$ such that $a^k x_k + b^k y_k = d^k$. We also have $x,y \in \mathbb{Z}$ such that $ax+by = d$. Hence, we get that $$(ax+by) \left( a^k x_k + b^k y_k \right)^2 = d^{2k+1}.$$
Expanding this, we get that $$a^{k+1} \left( a^k x_k^2 x + a^{k-1} b x_k^2 y + 2 b^k x_k y_k x \right) + b^{k+1} \left( b^k y_k^2 y + ab^{k-1} y_k^2 x + 2 a^k x_k y_k y \right) = d^{2k+1}.$$ Since $d|a$ and $d|b$, we have that $a = d e$ and $b = df$. Hence, we get that $$a^{k+1} \left( d^k e^k x_k^2 x + d^k e^{k-1} f x_k^2 y + 2 d^k f^k x_k y_k x \right) + b^{k+1} \left( d^k f^k y_k^2 y + d^k ef^{k-1} y_k^2 x + 2 d^k e^k x_k y_k y \right) = d^{2k+1}.$$
This gives us that
$$a^{k+1} \left( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x \right) + b^{k+1} \left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right) = d^{k+1}.$$
Hence, we have found integers $$x_{k+1} = \left( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x \right), \, y_{k+1} = \left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right)$$ such that $$a^{k+1} x_{k+1} + b^{k+1} y_{k+1} = d^{k+1}.$$
Hence, we have that $\gcd(a^{k+1}, b^{k+1}) \vert d^{k+1}$. It is also true that $d^{k+1} \vert a^{k+1}$ and $d^{k+1} \vert b^{k+1}$, since $d \vert a$ and $d \vert b$. Hence, $d^{k+1} \vert \gcd(a^{k+1}, b^{k+1})$. Hence, we get that $$\gcd(a^{k+1}, b^{k+1}) = \gcd(a,b)^{k+1}.$$
Hence, by the principle of mathematical induction, we have that $$\gcd(a^n,b^n) = \gcd(a,b)^n, \,\, \forall n \in \mathbb{Z}^+.$$
