$(a^{n},b^{n})=(a,b)^{n}$ 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 $d\mid b^n$

Show that the common divisors of $a^n$ and $b^n$ all divide $(a,b)^n$ and that any divisor of $(a,b)^n$ divides $a^n$ and $b^n$ (the proofs are pretty straight forward). It might be useful to consider prime factorization for the second direction. 
Similarly, show that $[a,b]^n$ is a multiple of $a^n$ and $b^n$ but that it is also the smallest such multiple (and again there, prime factorization might be useful in the second case). If you need more details just ask.
EDIT : 
Let's use prime factorization, I think this way everything makes more sense. To show equality, it suffices to show that the prime powers dividing either side are the same. Let's show $(a^n, b^n) = (a,b)^n$ first. 
Since $p$ is prime, $p$ divides $a$ if and only if it divides $a^n$ and similarly for $b$ ; so if $p$ does not divide $a$ or $b$, then $p^0 = 1$ is the greatest power of $p$ that divides both sides. If $p$ divides both $a$ and $b$, let $p^k$ be the greatest power of $p$ dividing $(a,b)$, so that $p^{kn}$ is the greatest power of $p$ dividing $(a,b)^n$. Since $p^k$ divides $a$, $p^{kn}$ divides $a^n$, and similarly for $b$. For obvious reasons the greatest power of $p$ dividing both $a^n$ and $b^n$ must be a power of $p^n$. But if $p^{(k+1)n}$ divided both $a^n$ and $b^n$, then $p^{(k+1)}$ would divide $a$ and $b$, contradicting the fact that $p^k$ is the greatest power of $p$ dividing $(a,b)$. Therefore $p^{kn}$ is the greatest power of $p$ dividing $(a^n,b^n)$ and the greatest power of $p$ dividing $(a,b)^n$, so taking the product over all primes, $(a^n,b^n) = (a,b)^n$. 
For $[a^n,b^n] = [a,b]^n$ you can do very similar techniques as with the gcd, except all the 'greatest' are replaced by 'smallest' in the last proof, and 'division' is replaced by 'being a multiple of'. 
Hope that helps,
