TITLE: On the vanishing of the generalized von Mangoldt function $\Lambda_k(n)$ when $n$ has more than $k$ prime factors
QUESTION [14 upvotes]: It is a well-known fact that the generalized von Mangoldt function, defined by
$$\displaystyle \Lambda_k(n) = \sum_{d | n} \mu(d) \left(\log \frac{n}{d}\right)^k$$
vanishes whenever $n$ has more than $k$ distinct prime factors. I was able to prove this fact through a relatively lengthy and cumbersome combinatorial argument by proving first for the squarefree case, say when $n = p_1 \cdots p_s$ with $s > k$, and showed that the vanishing of $\Lambda_k(n)$ can be deduced from the following polynomial identity:
$$\displaystyle (x_1 + \cdots + x_s)^k - \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-1}} \left(\sum_{i \in S} x_i\right)^k + \sum_{\substack{S \subset \{1, \cdots, s\} \\ |S| = s-2}} \left(\sum_{i \in S} x_i \right)^k - \cdots $$
$$ + (-1)^{s-1}\sum_{i=1}^s x_i^k = 0.$$
The general case is then done through a similar argument (which depends on the above identity) and strong induction. 
However, I find the above argument to be somewhat insipid and not very 'number theoretic', as it is deduced from a general polynomial identity rather than using any properties of numbers. Is there any conceptually simpler, more number theoretic proof? It would be an added bonus if the proof is shorter than the above.

REPLY [2 votes]: It can be seen as a consequence of the multinomial theorem. We would like to prove the following.
The function $\Lambda_k(n)$ is always non-negative and supported only on integers $n$ with $\omega(n)\leq k$, with $\omega(n)$ the distinct prime factors counting function. Moreover, it verifies the following recursive relation
$$
\Lambda_{k+1}(n)=\Lambda_k(n)\log n+\sum_{d|n}\Lambda_k(d)\Lambda(n/d).
$$
Proof:
If $\mu(n)$ indicates the Moebius function, we begin with the following series of identities
\begin{align*}
\sum_{\substack{b|n}}\mu(b)(\log b)^k &=\sum_{\substack{b\geq 1\\ b|n}}\frac{\mu(b)(\log b)^k}{b^{\sigma}}\bigg|_{\sigma=0}\\
&=(-1)^{k}\frac{d^{k}}{d\sigma^{k}}\sum_{\substack{b\geq 1\\ b|n}}\frac{\mu(b)}{b^{\sigma}}\bigg|_{\sigma=0}\\
&=(-1)^{k}\frac{d^{k}}{d\sigma^{k}}\bigg(\prod_{p|n}\bigg(1-\frac{1}{p^{\sigma}}\bigg)\bigg)\bigg|_{\sigma=0}\\
&=(-1)^{k}\sum_{j_1+j_2+...+j_{\omega(n)}=k}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}\bigg(1-\frac{1}{p_i^{\sigma}}\bigg)^{(j_i)}\bigg|_{\sigma=0},
\end{align*}
by the multinomial theorem. By differentiating each binomial above, we can rewrite the previous sum as
\begin{align*}
&=(-1)^{k+\omega(n)}\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(-\log p_i)^{j_i}.\\
&=(-1)^{\omega(n)}\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(\log p_i)^{j_i}.
\end{align*}
Moreover, we have 
$$\Lambda_k(n)=\sum_{b|n}\mu(n/b)(\log b)^k=\mu(n)\sum_{b|n}\mu(b)(\log b)^k=(-1)^{\omega(n)}\sum_{b|n}\mu(b)(\log b)^k.$$
We thus deduce that
$$\Lambda_k(n)=\sum_{\substack{j_1+j_2+...+j_{\omega(n)}=k\\ j_i\neq 0,\ \forall i}}\binom{k}{j_1,j_2,...,j_{\omega(n)}}\prod_{i=1}^{\omega(n)}(\log p_i)^{j_i},$$
from which it is immediate to obtain the first assertion. Regarding the second one, we first notice that the Dirichlet series of $\Lambda_k(n)$ is given by
$$(*)\ 
\sum_{n\geq 1}\frac{\Lambda_k(n)}{n^s}=(-1)^k\frac{\zeta^{(k)}(s)}{\zeta(s)}\ \ \ (\Re(s)>1),
$$
where $\zeta(s)$ is the Riemann zeta function. This follows immediately by looking at $\Lambda_k(n)$ as the Dirichlet convolution of the Moebius function and the logarithm function at $n$ and using some basic identities in the theory of Dirichlet series.  
Moreover, we clearly have
$$(-1)^{k}\sum_{n\geq 1}\frac{\Lambda_{k-1}(n)\log n}{n^s}=\frac{d}{ds}\frac{\zeta^{(k-1)}(s)}{\zeta(s)}=\frac{\zeta^{(k)}(s)}{\zeta(s)}-\frac{\zeta^{'}(s)}{\zeta(s)}\frac{\zeta^{(k-1)}(s)}{\zeta(s)}.$$
Plugging the relation (*), for $k$, $k-1$ and $1$, into the above, we get also the second assertion.