$(a^{n},b^{n})=(a,b)^{n}$ and $[a^{n},b^{n}]=[a,b]^{n}$? How to show that $$(a^{n},b^{n})=(a,b)^{n}$$ and $$[a^{n},b^{n}]=[a,b]^{n}$$ without using modular arithmetic? Seems to have very interesting applications.$$$$Try: $(a^{n},b^{n})=d\Longrightarrow d\mid a^{n}$ and $d\mid b^n$

let gcd(a,b)=d 
$\Longrightarrow d|a \wedge d|b$ 
$\Longrightarrow$ $d|a^n$ $\wedge$ $d|b^n$
$\Longrightarrow$ d|$(a^n,b^n)$---------------(1)
now, $d|(a,b)$     [in fact, $(a,b)=d$]
it implies $d|(a,b)^n$-----------------(2)
by equation (1)and(2), it follows obviously that $(a^n,b^n)=(a,b)^n$
now, it is easy work to derive similar result for lcm by this method.
