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
$$
\mu\left(n\right)=\begin{cases}
1 & \mbox{if }n=1\\
\left(-1\right)^{t} & \mbox{if }n\mbox{ is a product of }t\mbox{ distinct primes}\\
0 & \mbox{if }p^{2}\mbox{ divides }n\mbox{ for some prime }p.
\end{cases}
$$

$$ \forall n \geq 1 ,  g(n) = \sum_{d \mid n} f(d) \Rightarrow f(n) = \sum_{d \mid n} \mu(d) g \left(\frac{n}{d} \right) $$
$$ n = \sum_{d \mid n}\varphi(d) \Rightarrow \varphi(n) = \sum_{d \mid n} \frac{n\mu(d)}{d} = \sum_{d \mid n} d\mu \left( \frac{n}{d} \right) $$ 
proof of Möbius inversion formula is here http://www.proofwiki.org/wiki/M%C3%B6bius_Inversion_Formula 
