# Is there a combinatorial proof that Euler's totient function divides Jordan's totient function?

Jordan's totient function $$J_{k}(n)$$ is a generalization of Euler's totient function that counts the number of $$k$$-tuples $$(a_1, \ldots, a_k)$$ for which $$1 \leq a_1, \ldots, a_n \leq n$$ and $$gcd(a_1, \ldots, a_k, n) = 1$$, where $$n$$ and $$k$$ are positive integers. There is an explicit formula for this function as follows: $$J_{k}(n) = n^k \prod_{p} \left(1 - \frac{1}{p^k}\right),$$ where $$p$$ ranges over through the prime divisors of $$n$$. Using this identity, it is pretty straightforward to deduce that $$J_{1}(n)$$, which is Euler's totient function, divides $$J_{k}(n)$$ for all positive integers $$k$$ and $$n$$. However, given the simplicity of the definitions, I believe there is a much more elegant combinatorial proof of this fact that does not rely on the explicit formula. I would be glad if anyone could provide a combinatorial proof or trick that I'm unable to see.

Equivalently, $$J_k(n)$$ counts the $$k$$-tuples $$(a_1,\cdots,a_k)$$ of elements of $$\mathbb{Z}/n\mathbb{Z}$$ which generate the whole ring (as an ideal). We can verify $$(\mathbb{Z}/n\mathbb{Z})^\times$$ acts freely on these tuples by multiplication: if $$(a_1,\cdots,a_k)$$ is fixed under multiplication by $$u$$, and $$r_1a_1+\cdots+r_ka_k=1$$, then
$$u = u(r_1a_1+\cdots+r_ka_k)=r_1(ua_1)+\cdots+r_k(ua_k)=r_1a_1+\cdots+r_ka_k=1,$$
i.e. $$u=1$$. Since $$\varphi(n)=|(\mathbb{Z}/n\mathbb{Z})^\times|$$, this shows $$\varphi(n)\mid J_k(n)$$.
What I don't see is why $$d\mid k$$ would imply $$J_d(n)\mid J_k(n)$$ in general, though.