$\gcd(f,g)=1 \implies \gcd(f^n,g^n) = 1$ Given $f,g\in \Bbb F[x]$ such that $\gcd(f,g)=1$,how to prove $\gcd(f^n,g^n)=1$,for $n=1,2,\ldots$?
It seems quite obvious but I can't figure out a formal proof...
 A: Suppose $p$ is a prime that divides
$$\gcd(f^n,g^m)$$
then $p$ divides $f^n$ and hence divides $f$ (easily established using induction). Similarly for $g^m$. Since $\gcd(f,g)=1$, no such $p$ can exist.
Basic fact: If $p$ divides $f^n, n>1$ than $p$ divides $f$.
Proof: Clearly true for $n=1$.
Assume it true for $n$. Then for $n+1$
$p$ divides $f^{n+1}=f\cdot f^n$ implies that $p$ divides $f$ or it divides $f^n$. By hypothesis, $p$ divides $f$
Note: My result does not require the powers of $f$ and $g$ to be the same. Also, $n=0$ or $m=0$ is trivially valid
A: Since $f$ and $g$ are relatively prime, by Bezout's Identity there exist polynomials $p$ and $q$ such that $fp+gq=1$. Take the $2n-1$-th power of both sides. Any term of the binomial expansion of $(fp+gq)^{2n-1}$ is divisible by one of $f^n$ or $g^n$. It follows that
$$1=(fp+gq)^{2n-1}=f^n P+g^nQ$$
for some polynomials $P$ and $Q$. It follows that $f^n$ and $g^n$ are relatively prime.
A: More generally, for any ideals $I, J$ in a commutative ring $R$ with $1$, $\sqrt{I+J} = \sqrt{\sqrt{I}+\sqrt{J}}$, where $\sqrt{}$ denotes radical. Thus $I, J$ are coprime iff $I+J = R$ iff $\sqrt{I+J} = R$ iff $\sqrt{I}+\sqrt{J} = R$ iff $\sqrt{I}, \sqrt{J}$ are coprime. 
Applying this with $I = (f^n)$ and $J = (g^n)$ gives $(f^n, g^n) = 1$ iff $(f,g) = 1$, since $\sqrt{(f)} = \sqrt{(f^n})$.
