A proof of $\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}$ I'm trying to proof the following statement: 
Let $n \in \mathbb{Z}$ and the $\sum$ are on the divisors $d$ of $n$.  Show that
$$\sum\limits_{d|n} \sigma(d) = n \sum\limits_{d|n} {\tau(d) \over d}.$$
I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?

EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):

We have $$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{t\mid d}t=\sum_{t\mid n}\sum_{k\mid \frac{n}{t}}t=\sum_{t\mid n}t\sum_{k\mid \frac{n}t}1,$$
  where $k=\frac{d}{t}$.  So
  $$\sum_{d\mid n}\sigma(d)=\sum_{t\mid n}t\tau\left(\frac{n}t\right)=\sum_{\delta \mid n}\frac{n}{\delta}\tau(\delta)=n\sum_{\delta\mid n}\frac{\tau(\delta)}{\delta},$$
  where $\delta =\frac{n}{t}$.

 A: We have $$\sum_{d\mid n}\sigma(d)=\sum_{d\mid n}\sum_{t\mid d}t=\sum_{t\mid n}\sum_{k\mid \frac{n}{t}}t=\sum_{t\mid n}t\sum_{k\mid \frac{n}t}1,$$
where $k=\frac{d}{t}$.  So
$$\sum_{d\mid n}\sigma(d)=\sum_{t\mid n}t\tau\left(\frac{n}t\right)=\sum_{\delta \mid n}\frac{n}{\delta}\tau(\delta)=n\sum_{\delta\mid n}\frac{\tau(\delta)}{\delta},$$
where $\delta =\frac{n}{t}$.
A: Here  is a derivation with some intermediate  steps which might be helpful.

We obtain
  \begin{align*}
\color{blue}{\sum_{d\mid   n}\sigma(d)}&=\sum_{d\mid n}\sum_{t\mid d}t\tag{1}\\
&=\sum_{{t\mid d\mid n}\atop{t,d\geq 1}}t\tag{2}\\
&=\sum_{{t \mid tk \mid n}\atop{t,k\geq 1}}t\tag{3}\\
&=\sum_{{t \mid n,k \mid \frac{n}{t}}\atop{t,k\geq 1}}t\tag{4}\\
&\,\,\color{blue}{=\sum_{t \mid n}t\sum_{k \mid \frac{n}{t}}1}\tag{5}
\end{align*}

Comment:


*

*In (1) we use the definition of the divisor function $\sigma(d)=\sum_{t| d}t$.

*In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of  the summands.

*In (3) we use $t|d  \Longleftrightarrow  \exists k\geq 1 :  tk=d$, assuming $t,d,k$ are positive integers.

*In (4) we use the transitivity of the $|$ operator: $t| d| n\Longrightarrow t|n$ and we also use $t \cdot k| n\Longleftrightarrow t | \frac{n}{k}$.

*In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$  from  the  inner  sum, since $t$ does not  depend on the index  variable  $k$.

[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all  $n$ and the function $N(n)=n$ for all $n$ we have 
  \begin{align*}
\tau(n)&=\sum_{d|n}1=(u\star u)(n)\\
\sigma(n)&=\sum_{d|n}d=(u\star N)(n)
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\sum_{d|n}\sigma(d)}&=(\sigma  \star u)(n)\\
&=((u\star N)\star u)(n)\\
&=(u\star(N \star u))(n)\\
&=(u\star (u\star N))(n)\\
&=((u\star u) \star N)(n)\\
&=(\tau \star N)(n)\\
&\,\,\color{blue}{=\sum_{d|n}\tau(d)\frac{n}{d}}
\end{align*}
  and the claim follows.

