Arithmetical function : How can I prove? How can I show that this sum $\sum_{d|n} \mu(d) \log^kd$ is $0$ where $\mu(d)$ is mobius function.
I've expect that this question is solved by induction..! 
$k$ is integer that is a power of $\log$

For $k=1$ the identity should be the following
$$S_1(n)=\sum_{d|n}\mu(d)\ln(d)=\begin{cases}
0& \mbox{if $n$ is not a power of a prime},\\
-\ln(p)&\mbox{if $n$ is a power of a prime $p$.}
\end{cases}$$
Let $n=p_1^{a_1}\cdots p_r^{a_r}$ with $r>0$ then
$$S_1(n)=-\sum_{1\leq i_1\leq r}\ln(p_{i_1})+\sum_{1\leq i_1<i_2\leq r}(\ln(p_{i_1})+\ln(p_{i_2}))-\sum_{1\leq i_1<i_2<i_3\leq r}(\ln(p_{i_1})+\ln(p_{i_2})+\ln(p_{i_3}))+\cdots+ (-1)^r(\ln(p_{i_1})+\dots+\ln(p_{i_r})).$$
Now we count the terms $\ln(p_i)$ in the above sum
$$-1+(r-1)-\frac{1}{2}(r-1)(r-2)+\dots +(-1)^r=-\sum_{j=0}^{r-1} (-1)^{j}\binom{r-1}{j}\\=-(1-1)^{r-1}=\begin{cases}
0& \mbox{if $r>1$},\\
-1&\mbox{if $r=1$.}
\end{cases}$$
and we are done.
