Seeking for a proof on the relation between Euler totient and Möbius function Can someone help me prove the relation
$\varphi\left(n\right)={\displaystyle \sum_{d|n}}d\mu\left(n/d\right)$,
where $\mu$ is the Möbius function defined by
$$
\mu\left(n\right)=\begin{cases}
1 & \mbox{if }n=1\\
\left(-1\right)^{t} & \mbox{if }n\mbox{ is a product of }t\mbox{ distinct primes}\\
0 & \mbox{if }p^{2}\mbox{ divides }n\mbox{ for some prime }p.
\end{cases}
$$

Hint: 
$$\varphi(p_1^{m_1}p_2^{m_2}\ldots p_k^{m_k}) = p_1^{m_1}p_2^{m_2}\ldots p_k^{m_k} \prod_{i=1}^k \left( 1 - \frac{1}{p_k}\right)$$ and $d \mid n$ iff $d=p_1^{r_1}p_2^{r_2}\ldots p_k^{r_k}$ for $0 \le r_i \le m_i$.
