How to use fundamental theorem of arithmetic to conclude that $\gcd(a^k,b^n)=1$ for all $k, n \in$ N whenever $a,b \in$ N with $\gcd(a,b)=1$? How to use fundamental theorem of arithmetic to conclude that $\gcd(a^k,b^n)=1$ for all $k, n \in$ N whenever $a,b \in$ N with $\gcd(a,b)=1$?
Fundamental theorem of arithmetic: Each number $n\geq 2$ can be presented in an unique way as a product
$n=p_1^{a_1}p_2^{a_2} \dots p_r^{a_r}$, where
$p_i$ are primes $p_1<p_2< \dots <p_r$ and $a_i \in$ N.
 A: The prime factors of $a^k$ are exactly the same as the prime factors of $a$.
Hence, a prime $p$ divides both $a^k$ and $b^n$ iff it divides both $a$ and $b$. Furthermore, $\gcd(a,b)=1$ iff $a$ and $b$ share no prime factors. Thus $\gcd(a^k,b^n)=1$ iff $\gcd(a,b)=1$.
A: Hint:  $\gcd(a,b)=1$ if and only if $a$ and $b$ share no prime factors.
A: Hint $\rm\ $ prime $\rm\ p\: |\: n^k\ \Rightarrow\ p\: |\: n\ $ by uniqueness of prime factorizations.
Note $\ $ In fact uniqueness of factorizations into primes (i.e. atoms) is equivalent to the following  
Prime Divisor Property  $\rm\quad p\ |\ a\:b\ \Rightarrow\ p\:|\:a\ $ or $\rm\ p\:|\:b,\ $ for all primes $\rm\:p\,;\:\: $ more generally
Primal Divisor Property  $\rm\ \ \: c\ |\ a\:b\ \Rightarrow\ c_1\, |\: a\:,\: $ $\rm\ c_2\:|\:b,\ \ c = c_1\:c_2,\ $ for all $\rm\:c$
The latter property may be considered to be a generalization of the prime divisor property from atoms to composites (one easily checks that atoms are  primal $\Leftrightarrow$ prime). This leads to various "refinement" views of unique factorizations, e.g. via Schreier refinement and Riesz interpolation, the Euclid-Euler Four Number Theorem (Vierzahlensatz), etc, which prove more natural in noncommutative rings - see  Paul Cohn's 1973 Monthly survey Unique Factorization Domains.
A: One way to go about this is to first prove that $(a,b) = 1$ and $(a,c) =1$ implies $(a, bc)=1$. Your problem then follows by induction.
To prove the lemma, notice that $ax + by = 1$ and $az + cw = 1$ imply
$$ax + by(az+cw) = 1$$
$$ax + byaz + (bc)yw = 1$$
$$a(x+byz) + bc (yw) = 1.$$
