Greatest common divisors equal? Let $a,b$ be natural numbers. Show that $gcd(a^n,b^n)$ = ($gcd(a,b)^n)$ for any integer $n$.
How I started was first proof by contradiction, and then tried to do an inductive proof when that didn't work, but neither of them worked out for me. I think I'm confused as to the notation I should use. Hmmm.
I'm not looking for someone to answer the question fully, I just need a nudge in the right direction.
EDIT:
Okay, after the super helpful hints, here is what I have for the proof:
Suppose $gcd(a,b)$ = $c$ for some integer $c$ 
Then $c$ = ${p_1}^{min\{a_1, b_1\}\ }$... ${p_k}^{min\{a_k, b_k\}}$ for some positive integer $k$
So $gcd(a^n, b^n)$ = ${p_1}^{min\{na_1, nb_1\}\ }$... ${p_k}^{min\{na_k, nb_k\}}$ = ${p_1}^{nmin\{a_1, b_1\}\ }$... ${p_k}^{nmin\{a_k, b_k\}}$
Now we see ($gcd(a, b)^n$) = $(c)^n$ = (${p_1}^{min\{a_1, b_1\}\ }$... ${p_k}^{min\{a_k, b_k\}})^n$ = ${p_1}^{nmin\{a_1, b_1\}\ }$... ${p_k}^{nmin\{a_k, b_k\}}$
Therefore, $gcd(a^n, b^n)$ = $(gcd(a, b)^n)$ for all positive integers $n$
Is this too short? Should I show the prime factorization of $a$ and $b$ before showing the prime factorization of $gcd(a, b)$ ?

A start: Let $d=\gcd(a,b)$, and let $a=da_1$ and $b=db_1$. We want to show that $\gcd(a^n,b^n)=d^n$, or equivalently that $\gcd(a_1^n,b_1^n)=1$. 
