Show that $\sigma(n) = \sum_{d|n} \phi(n) d(\frac{n}{d})$ This is a homework question and I am to show that $$\sigma(n) = \sum_{d|n} \phi(n) d\left(\frac{n}{d}\right)$$ where $\sigma(n) = \sum_{d|n}d$, $d(n) = \sum_{d|n} 1 $ and $\phi$ is the Euler Phi function. 
What I have. Well I know $$\sum_{d|n}\phi(d) = n$$ I also know that for $n\in \mathbb{Z}^n$ it has a certain prime factorization $n = p_1^{a_1} \ldots p_k^{a_k}$ so since $\sigma$ is a multiplicative function, we have $\sigma(n) = \sigma(p_1)\sigma(p_2)...$ 
I also know the theorem of Möbius Inversion Formula and the fact that if $f$ and $g$ are artihmetic functions, then $$f(n) = \sum_{d|n}g(d)$$ iff $$g(n) = \sum_{d|n}f(d)\mu\left(\frac{n}{d}\right)$$
Please post no solution, only hints. I will post the solution myself for others when I have figured it out.

First hint: verify the formula when $n$ is a power of a prime. Then, prove that the function $n \mapsto \sum_{d \mid n} \phi(n) d(n/d)$ is also multiplicative, so it must coincide with $\sigma$. In fact, prove more generally that if $a_n$ and $b_n$ are multiplicative, then $n\mapsto \sum_{d \mid n}a_db_{n/d}$ is multiplicative.
Second hint (more advanced): If $A(s) = \sum_{n\geq 1}\frac{a_n}{n^{-s}}$ and $B(s) = \sum_{n\geq 1}\frac{b_n}{n^{-s}}$ are Dirichlet series, their product is $$A(s)B(s)=\sum_{n\geq 1}\frac{(a * b)_n}{n^{-s}},$$
where $(a*b)_n = \sum_{d\mid n} a_n b_{n/d}$. Express the Dirichlet series $\sum_{n\geq 1}\frac{\phi(n)}{n^{-s}}$, $\sum_{n\geq 1}\frac{d(n)}{n^{-s}}$ and $\sum_{n\geq 1}\frac{\sigma(n)}{n^{-s}}$ in terms of the Riemann zeta function, and reduce your identity to an identity between them.
