A divisor sum involving generalized Moebius function I am trying to prove the following claim:

Let $n$ be a squarefree natural number. Denote by $\mu_k$ the generalized Moebius function: $\mu_k=\underbrace{\mu \ast \ldots \ast \mu}_{k}$ where $\ast$ is a Dirichlet convolution, and let $\omega(n)$ be the prime omega function. Then,
$$\displaystyle\sum_{d \mid n} |\mu_k(d)|=(k+1)^{\omega(n)} \quad \text{for} \quad k\ge 1$$

The SageMath cell that demonstrates this claim can be found here.
My attempt:
I tried to apply a mathematical induction:

*

*Induction base (k=1): $\displaystyle\sum_{d \mid n} |\mu(d)|=2^{\omega(n)}$ - This is a well known identity.


*Induction hypothesis (k=m): $\displaystyle\sum_{d \mid n} |\mu_m(d)|=(m+1)^{\omega(n)}$


*Induction step (k=m+1): $\displaystyle\sum_{d \mid n} |\mu_{m+1}(d)|=(m+2)^{\omega(n)}$
Since $\mu_{m+1}=\mu_m \ast \mu$ we have:
$$\displaystyle\sum_{d \mid n} |(\mu_{m} \ast \mu)(d)|=(m+2)^{\omega(n)}$$
$$\displaystyle\sum_{d \mid n} \left|(\mu_{m}(d)\mu \left(\frac{n}{d}\right)\right|=(m+2)^{\omega(n)}$$
Now, I am stuck here. How should I proceed?
 A: First, your last goal is incorrectly derived from the penultimate one: it should read
$$
\sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c)\mu \biggl(\frac{d}{c}\biggr)\biggr|=(m+2)^{\omega(n)}.
$$
Now note that since $n$ is squarefree, $d$ and $d/c$ must also be squarefree, and so $|\mu(d/c)=1|$ always. Using the induction hypothesis and sorting the left-hand side according to the value of $\omega(d)$ then yields
\begin{align*}
\sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c)\mu \biggl(\frac{d}{c}\biggr)\biggr| &= \sum_{d \mid n} \biggl| \sum_{c\mid d} \mu_{m}(c) \biggr| \\
&= \sum_{d \mid n} (m+1)^{\omega(d)} \\
&= \sum_{k=0}^{\omega(n)} (m+1)^k \sum_{\substack{d \mid n \\ \omega(d)=k}} 1 \\
&= \sum_{k=0}^{\omega(n)} (m+1)^k \binom{\omega(n)}k = \big((m+1)+1)^{\omega(n)}=(m+2)^{\omega(n)}
\end{align*}
by the binomial theorem. (The fact that $\#\{d\mid n\colon \omega(d)=k\} = \binom{\omega(n)}k$ is because such divisors $d$ correspond exactly to $k$-element subsets of the set of $\omega(n)$ prime factors of $n$.)
