On the vanishing of the generalized von Mangoldt function $\Lambda_k(n)$ when $n$ has more than $k$ prime factors

Everything I know about these generalized von Mangoldt functions goes through the identity $$ \Lambda_{k+1}(n) = (\Lambda_k \ast \Lambda)(n) + \Lambda_k(n) \log n $$ (where $\ast$ denotes Dirichlet convolution); this identity isn't obvious, but can be proved by identifying the Dirichlet series corresponding to both sides. But in particular, the desired statement about the support of $\Lambda_k$ follows easily by induction from this recursive identity.

Write the Riemann zeta function as a product of its Euler factors $\zeta (s)=\prod_i E_{i}(s)$. Repeated application of the Leibniz rule shows $$\frac{\zeta^{(k)}(s)}{\zeta (s)}=\sum_{i_1+\cdots+i_k=k}\sum_{t_1,\dots,t_k}\frac{E_{t_1}^{(i_1)}(s)}{E_{t_1}(s)}\cdots \frac{E_{t_k}^{(i_k)}(s)}{E_{t_k}(s)}.$$ The left hand side is clearly the Dirichlet series for the generalized Von Mangoldt function, and the right hand side is a Dirichlet series that is supported on integers with at most $k$ prime factors.

This is a supplement to the responses so far. Here is a proof avoiding Dirichlet series that $$ \Lambda_{k+1}=\Lambda_k\log+\Lambda\ast\Lambda_k. $$ By definition, $$ \Lambda_{k+1}=\mu\ast(\log^k\log)=(\mu\ast\log^k)\log+(-\mu\log)\ast\log^k $$ $$ =(\mu\ast\log^k)\log+(-\mu\log)\ast 1\ast\Lambda_k =\Lambda_k\log+\Lambda\ast\Lambda_k. $$ In the first line, we used that $\log(n/d)=\log(n)-\log(d)$. In the second line, we used that $\Lambda=(-\mu\log)\ast 1$, which is the special case of the first line for $k=0$.

P.S. From this identity one can prove by induction on $k$ that $\Lambda_k(n)\geq 0$, with equality iff $n$ has more than $k$ distinct prime factors.