Define $\Lambda_k(n,x)=\sum\limits_{d|n}\mu(d)\left(\ln\frac xd\right)^k$, where $n$ is a positive integer and $x$ is a positive real number.
Let $n=p_1^{a_1}...p_r^{a_r}$, prove that $\Lambda_k(n,x)\leq r!\binom kr(\ln x)^{k-r}(\ln p_1)...(\ln p_r)$.
Remark. It's Eq(1.49) of Analytic Number Theory of Iwaniec.
Given your comment, I'll assume your comfortable with the following multiplication formula $$ \tag{1} \Lambda_k(mn,x) = \sum_{0\leq j\leq k} \binom{k}{j} \Lambda_j(n) \Lambda_{k-j}\left(m,\frac{x}{n}\right), \qquad (m,n)=1. $$ We wish to show that if $n=p_1^{a_1}\cdots p_r^{a_r}$, then $$ \tag{2} \Lambda_k(n,x) \leq r! \binom{k}{r}(\log x)^{k-r} (\log p_1)\cdots(\log p_r). $$ If $n=p^a$ (so $r=1$), we can just compute this directly: $$ \Lambda_k(p^a,x) = (\log x)^k - \left(\log\frac{x}{p}\right)^k = (\log x)^k \left(1-\left(1-\frac{\log p}{\log x}\right)^k\right) \leq (\log x)^k \left(\frac{k\log p}{\log x} \right) $$ upon using the inequality $$ 1-(1-t)^k \leq k t\quad \text{for $t\in[0,1]$}. $$ Thus we have (2) when $r=1$.
For the induction step, write $n=p_1^{a_1}\cdots p_{r-1}^{a_{r-1}}$ with $p_1<\cdots<p_{r-1}$ and $r\leq k$ and suppose that suppose that (2) holds for integers with $r-1$ prime factors. Let $p_r$ be a prime with $p_r>p_{r-1}$. Then since $\Lambda_k(p_r^a) = (\log p_r)^k (a^k-(a-1)^k)$, we have by (1) and the induction hypothesis that $$ \begin{aligned} \Lambda_k(p_r^a n,x) &= \sum_{j=0}^{k-r+1} \binom{k}{j} \Lambda_j(p_r^a) \Lambda_{k-j}\left(n,\frac{x}{p_r^a}\right) \\ &\leq \sum_{j=0}^{k-r+1} \binom{k}{j} \left[ (\log p_r)^j (a^j-(a-1)^j) \right]\\ &\qquad\qquad \times \left[(r-1)! \binom{k-j}{r-1} \left(\log \frac{x}{p_r^a}\right)^{k-j-(r-1)} (\log p_1)\cdots (\log p_{r-1}) \right]\\ &= (r-1)!(\log p_1)\cdots (\log p_r) \\ &\qquad\qquad\times\sum_{j=0}^{k-r+1} \binom{k}{j}\binom{k-j}{r-1} (a^j-(a-1)^j) (\log p_r)^{j-1}\left(\log \frac{x}{p_r^a}\right)^{k-j-(r-1)}. \end{aligned} $$ Since (see derivation below) $$ \binom{k}{j}\binom{k-j}{r-1} = \frac{r}{j}\binom{k}{r}\binom{k-r}{j-1} , $$ and the summand above is $0$ when $j=0$, the last line above is $$ r! \binom{k}{r} (\log p_1)\cdots (\log p_r) \sum_{j=1}^{k-r+1} \frac{1}{j}\binom{k-r}{j-1} (a^j-(a-1)^j)(\log p_r)^{j-1} \left(\log \frac{x}{p_r^a}\right)^{k-r-(j-1)}. $$ The sum over $j$ is less than or equal to $$ \begin{aligned} \sum_{j=1}^{k-r+1} \binom{k-r}{j-1} a^{j-1} (\log p_r)^{j-1}\left(\log \frac{x}{p_r^a}\right)^{k-r-(j-1)} &= \sum_{j=0}^{k-r} \binom{k-r}{j} (\log p_r^a)^j \left(\log \frac{x}{p_r^a}\right)^{k-r-j} \\ &= \left(\log \frac{x}{p_r^a}+\log p_r^a\right)^{k-r} \\ &= (\log x)^{k-r}. \end{aligned} $$ Derivation of binomial identity:
$$ \begin{aligned} \binom{k}{j}\binom{k-j}{r-1} &= \frac{k!}{j!(k-j)!} \cdot \frac{(k-j)!}{(r-1)!(k-j-(r-1))!} \\ &= r \frac{k!}{r!(k-r)!} \cdot\frac{(k-r)!}{j!(k-j-(r-1))!}\\ &=\frac{r}{j} \frac{k!}{r!(k-r)!} \cdot\frac{(k-r)!}{(j-1)!(k-r-(j-1))!}\\ &=\frac{r}{j} \binom{k}{r}\binom{k-r}{j-1}. \end{aligned} $$