A divisor sum involving Moebius function and Jordan's totient function I am trying to prove the following claim:

Let $\mu(n)$ be the Moebius function and let $J_k(n)$ be the Jordan's totient function. Then,
$$\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{J(k,d)}=\frac{n^k}{J(k,n)} \quad \text{for} \quad k\ge 1$$

The SageMath cell that demonstrates this claim can be found here.
My attempt:
For proof method I have chosen mathematical induction:

*

*Induction base (k=1): $\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}=\frac{n}{\varphi(n)}$ - This is proven identity (see here).


*Induction hypothesis (k=m): $\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{J(m,d)}=\frac{n^m}{J(m,n)}$


*Induction step (k=m+1): $\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{J(m+1,d)}=\frac{n^{m+1}}{J(m+1,n)}$
By applying definition of Jordan's totient function we have:
$$\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{J(m+1,d)}=$$$$\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{d^{m+1} \displaystyle\prod_{p \mid d} \left(1-\frac{1}{p^{m+1}}\right)}=$$$$\displaystyle\sum_{d \mid n} \frac{\mu^2(d)}{d \cdot d^{m} \displaystyle\prod_{p \mid d} \left(1-\frac{1}{p \cdot p^{m}}\right)}$$
How should I proceed?
 A: As pointed out in the comments, your intuition should probably not be pointing at induction here, because the identities for $k$ and $k+1$ seem unconnected.

An arithmetic function is multiplicative if $f(ab)=f(a)f(b)$ for all coprime whole numbers $a$ and $b$ (if it is true for all whole numbers with no restrictions, it is completely multiplicative). It is instructive to show that if $f$ is multiplicative, then so too is the function $F(n):=\sum_{d\mid n}f(d)$. Indeed, more generally, if $f$ and $g$ are multiplicative then so too is their convolution $(f\ast g)(n):=\sum_{d\mid n}f(d)g(n/d)$. I really recommend grappling with this fact to see why it is true.
Both $\mu(n)^2$ and $J_k(n)$ are multiplicative, so $\mu^2/J_k$ is multiplicative, so $\sum_{d\mid n}\mu(d)^2/J_k(d)$ is too. As is the arithmetic function $n^k/J_k(n)$ on the other side of the equation.
Another important fact about arithmetic functions is they are determined by their values on prime powers. Thus, to prove $F(n)=G(n)$ for all $n$, it suffices to prove $F(p^v)=G(p^v)$ for all prime powers $p^v$. Often this simplifies things. For instance, in our case the identity becomes
$$ \frac{1}{1}+\frac{1}{p^k(1-\frac{1}{p^k})}+0+0+\cdots=\frac{p^v}{J_k(p^v)} $$
which is readily verifiable, so the claim is proved.
