How to prove $\sum_{d|n} \mu(d) \sigma(n / d) = n$? How can I show that for all $n \in \mathbb{N}$;$$\sum_{d \mid n} \mu(d)\sigma(n/d) = n$$ 
Where $\sigma$ is the divisor function and $\mu$ is the Möbius function.
I am confused because isn't $\sum_{d \mid n}\mu(d)= 1$ if $n=1$ and 0 otherwise?

$$\sum_{d \mid n} d = \sigma(n)$$ and so by the Möbius inversion formula, we have that
$$\sum_{d \mid n} \mu(d) \sigma\left(\frac{n}{d}\right) = n$$
