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.