IX OBM

First observe that if $a_i\equiv b_i(\mathrm{mod}r)$, $i=1,2,\dots ,n$ then $p(a_1,a_2,\dots ,a_n)\equiv p(b_1,b_2,\dots ,b_n)(\mathrm{mod}r)$. If $u$ and $v$ are coprime then given $b_1,b_2,\dots ,b_n$ and $c_1,c_2,\dots ,c_n$ with $0\le b_i<u$ and $0\le c_i<v$ by the chinese remainder theorem there exist unique integers $a_1,a_2,\dots ,a_n$ with $0\le a_i<uv$ such that $a_i\equiv b_i(\mathrm{mod}u)$ and $a_i\equiv c_i(\mathrm{mod}v)$, which imply $p(a_1,a_2,\dots ,a_n)\equiv p(b_1,b_2,\dots ,b_n)(\mathrm{mod}u)$, $p(a_1,a_2,\dots ,a_n)\equiv p(c_1,c_2,\dots ,c_n)(\mathrm{mod}v)$. Thus $$\gcd (p(a_1,a_2,\dots ,a_n),uv)=1⟺\{\begin{array}{l}\gcd (p(a_1,a_2,\dots ,a_n),u)=1\\ \gcd (p(a_1,a_2,\dots ,a_n),v)=1\end{array}$$ and $k(uv)=k(u)\cdot k(v)$. For $p$ prime $\gcd (p(a_1,a_2,\dots ,a_n),p^s)=1⟺\gcd (p(a_1,a_2,\dots ,a_n),p)=1$. Divide each $a_i$ by $p$, obtaining quotient $q_i$ and remainder $r_i$. Reducing modulo $p$, we obtain $\gcd (p(a_1,a_2,\dots ,a_n),p)=1⟺\gcd (p(r_1,r_2,\dots ,r_n),p)=1$. Given $r_i$ there are $p^s/p=p^{s-1}$ possibilities for $q_i$ such that $0\le a_i<p^s$. Hence $k(p^s)=(p^{s-1}{)}^{n}k(p)=p^{n(s-1)}k(p)$, as required.