For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$ For any positive integer $n$, show that $\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d)$

My try :
Left hand side :
$\begin{align} \sum_{d|p^k}\sigma (d) &= \sigma(p^0) + \sigma(p^1) + \sigma(p^2) + \cdots + \sigma(p^k) \\ &= \dfrac{p^{0+1}-1}{p-1} +  \dfrac{p^{1+1}-1}{p-1}  + \cdots +  \dfrac{p^{k+1}-1}{p-1}  \\&= \dfrac{1}{p-1}\left( (p + p^2 + \cdots + p^{k+1}) - (k+1)\right) \\&= \dfrac{1}{p-1}\left(\dfrac{p(p^{k+1}-1)}{p-1}- (k+1)\right) \\
\end{align}$
I'm not sure if I am in right path; this doesn't look simple. Any help ?

looking at this intuitively, firstly we note that:
$$
\sum_{d|n}\sigma(d) = \sum_{d|n}(n/d)\tau(d) \\
=  \sum_{d|n} d\tau(n/d)
$$
so now we are summing the divisors $d$ of $n$, each divisor being counted with multiplicity $\tau(n/d)$. so you just have to persuade yourself that this multiplicity is appropriate.
but this is evident if we look at a particular $d$, since for  any integer  $m$ we have 
$$
dm | n  \Leftarrow \Rightarrow m | n/d
$$
in words there is a 1-1 correspondence between (a)  multiples of $d$ which divide $n$, and (b)  divisors of $n/d$ 
