$R$ is a integral domain, $p_1$ and $p_2$ are prime, $p_1$ and $p_2$ do not associate, $n_1,n_2 \ge 1,n_1,n_2 \in \mathbb N $, I need to show that $g:=\gcd({p_1}^{n_1},{p_2}^{n_2})$ associates with $1$.
I use induction for $n_1$. Below $r_1,r_2, ...$ are elements from $R$.
Assume that $n_1 =1$.
1) $gr_1=p_1,gr_2={p_2}^{n_2}$. Because $p_1$ prime, then $p_1 | r_1$ or $p_1 | g$.
2) If $p_1 | g$, then $p_1r_3r_2={p_2}^{n_2}$, thus $p_1$ associates with $p_2$, which is a contradiction.
3) Thus $p_1 | r_1 \implies p_1r_4=r_1 \implies gp_1r_4=p_1$ and it follows that $gr_4=1$, which means that $g$ associates with $1$.
Now $n_1 \ge 2$. The argumentation is the same up to step 3.
3') Thus $p_1 | r_1 \implies p_1r_4=r_1 \implies gp_1r_4={p_1}^{n_1}$ and it follows that $gr_4={p_1}^{n_1-1}$, which means that $g$ associates with $1$ because of the induction hypothesis.
I have two questions:
1) is the proof correct,
2) is there a more elegant proof which still uses basic facts about integral domains and prime elements?