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..

Here's a colored, edited version of user17762's answer. 
The proof is by induction. Let us denote $\gcd(a,b) = d$. It's given $\gcd(a,b) = 1   
\iff$
by Bezout's Lemma, there exist $x,y \in \mathbb{Z}$ such that $ax+by = d. (\dagger)$   
Assume that the statement is true for all $n \leq k$. Scilicet $\gcd(a^n,b^n) = d^n$ 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 $a^k x_k + b^k y_k = d^k$.    
Multiply this with $(\dagger)$ to result in $\begin{align} (ax+by) \left( a^k x_k + b^k y_k \right)^2 & = d^{2k+1}. \\
\color{green}{a^{2k + 1}xx_k^2 + a^{k + 1}2b^kx_ky_kx} \color{blue}{+ ab^{2k}yk^2x} &=  \\
+ \color{red}{a^{2k}bx_k^2y} + b^{k + 1}2a^kx_ky_ky + b^{2k+1}y_k^2y
\end{align}$
$LHS = \color{green}{a^{k+1} \left( a^k x_k^2 x \color{red}{+ 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 + \color{blue}{ab^{k-1} y_k^2 x} + 2 a^k x_k y_k y \right) $ 
Since $d|a$ and $d|b$, we have that $\color{brown}{a = d e}$ and $\color{brown}{b = df}$. $LHS = a^{k+1} \left( \color{brown}{d^k e^k} x_k^2 x + \color{brown}{d^k e^{k-1} f} x_k^2 y + 2\color{brown}{ d^k f^k} x_k y_k x \right) + b^{k+1} \left( \color{brown}{d^k f^k} y_k^2 y + \color{brown}{d^k ef^{k-1}} y_k^2 x + 2 \color{brown}{d^k e^k} x_k y_k y \right)$
Divide both sides by $d^k$ to result in 
$$a^{k+1} \underbrace{( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x)}_{\huge{x_{k+1}}} + b^{k+1} \underbrace{\left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right)}_{\huge{y_{k+1}}} = d^{k+1}.$$
Hence, we have found integers $x_{k+1}, \, y_{k+1}$ such that $a^{k+1} x_{k+1} + b^{k+1} y_{k+1} = d^{k+1}$
$ \iff \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, $d^{k+1} \vert \gcd(a^{k+1}, b^{k+1})$. 
In all , we get that $\gcd(a^{k+1}, b^{k+1}) = \gcd(a,b)^{k+1}.$
Hence, by the principle of mathematical induction $\gcd(a^n,b^n) = \gcd(a,b)^n, \,\, \forall n \in \mathbb{N}.$
