In class we saw a similar result when $k=2$, and now I'm trying to extend this to an arbitrary $k\in \mathbb{Z}^+$. When I plug in values this identity seems to hold, however I'm unsure how to tackle the proof.
This is how I started: Suppose that $p_{1},...,p_{q}$ are the distinct prime factors in the prime factorization of $n$ that have exponent $\geq 2$ in the prime factorization (it is possible that $q=0$. Other than when $d=1$, the only integers that will contribute to the sum are products of these primes with powers $\leq k-1$. Furthermore we see that any $d$ in which some exponent of a prime is $\geq k$ makes no contribution. I'm trying to write, $\sum_{d^k|n} \mu(d)=\sum_{d|p_{1}^{\alpha_{1}}\cdot...\cdot p_{q}^{\alpha_{q}}} \mu(d),$ where $\alpha_{i} \leq k-1$. This is where I get stuck and I'm not sure how to make a generalization about when this will evaluate to $0$ versus $1$. Any help or guidance would be greatly appreciated!
Let $(a,b)=1$, then
$$ \sum_{d^k|a}\mu(d)\sum_{r^k|b}\mu(r)=\sum_{(dr)^k|ab}\mu(dr)=\sum_{m|ab}\mu(m) $$
Hence, this function is multiplicative. As a result, let's consider prime powers $n=p^r$:
$$ \sum_{d^k|p^r}\mu(d)=\mu(1)+\mu(p)[k\le r] $$
If $k\le r$, the number $n$ is not $k$-free, so we conclude
$$ \sum_{d^k|n}\mu(d)=\begin{cases} 1 & \text{$n$ is $k$-free} \\ 0 & \text{otherwise} \end{cases} $$