Prove that $\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}$ Let $\omega(n)$ denote the number of distinct prime divisors of $n>1$, with $\omega(1)=0$. 
(a) Show that $2^{\omega(n)}$ is a multiplicative function. 
(b) Prove that $$\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}.$$
I have done the part (a) and I am stuck by (b). 
First, I set $d=p_1^{e_1}\cdots p_k^{e_{k}}$ be a factor of $n$. Then I don't know what's the next step.
 A: Suppose we seek to show that
$$\tau(n^2) =
\sum_{d|n} 2^{\omega(d)}.$$
This can be done using Dirichlet series and Euler products.
We have for the RHS and 
$$\sum_{n\ge 1} \frac{1}{n^s} 2^{\omega(n)}$$ 
the Euler product
$$\prod_p 
\left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}}
+\cdots\right).$$
which is
$$\prod_p \left(-1 + 2\frac{1}{1-1/p^s}\right)
= \prod_p \frac{-1+1/p^s+2}{1-1/p^s}
\\ = \prod_p \frac{1+1/p^s}{1-1/p^s}
= \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^2}
= \frac{\zeta(s)^2}{\zeta(2s)}.$$
Therefore 
$$\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} 2^{\omega(d)}
= \frac{\zeta(s)^3}{\zeta(2s)}.$$ 
On the other hand we have
$$\sum_{n\ge 1} \frac{1}{n^s} \tau(n^2)
\\= \prod_p 
\left(1 + (2+1) \frac{1}{p^s}
+ (4+1) \frac{1}{p^{2s}}
+ (6+1) \frac{1}{p^{3s}}
+ (8+1) \frac{1}{p^{4s}}
+ \cdots\right).$$
This is
$$\prod_p \left(1+\frac{1/p^s}{1-1/p^s}
+ \sum_{k\ge 1} \frac{2k}{p^{ks}}
\right)
\\ = \prod_p \left(1+\frac{1/p^s}{1-1/p^s}
+ 2 \frac{1/p^s}{(1-1/p^s)^2}
\right).$$
To aid in simplification we put $z=1/p^s$ to get
for the inner term
$$1 + \frac{z}{1-z}
+ \frac{2z}{(1-z)^2}$$
This simplifies to
$$\frac{1+z}{(1-z)^2}.$$
On the other hand
$$\frac{\zeta(s)^3}{\zeta(2s)}
= \prod_p \frac{1-z^2}{(1-z)^3}
= \prod_p \frac{1+z}{(1-z)^2}.$$
We have equality, QED.
A: A Combinatorial Proof (Sketch)
I assume that, by $\sigma$, you mean $\sigma_0=\tau$, the divisor-counting function.  Let $$S(n):=\Big\{(t,d)\in\mathbb{N}\times\mathbb{N}\,\Big|\,t\mid n^2,\,d\mid n,\text{ and }d\text{ is smallest such that }n^2\mid td^2\Big\}\,.$$
First, verify that, for each $t\in\mathbb{N}$ such that $t\mid n^2$, there exists a unique $d\in\mathbb{N}$ such that $(t,d)\in S(n)$.  This proves that $\big|S(n)\big|=\sigma_0\left(n^2\right)$.
Next, justify that, for each $d\in\mathbb{N}$ such that $d\mid n$, there are precisely $2^{\omega(d)}$ values of $t\in\mathbb{N}$ such that $(t,d)\in S(n)$.  This shows that $\big|S(n)\big|=\sum\limits_{d\mid n}\,2^{\omega(d)}$.  Therefore, $\sigma_0\left(n^2\right)=\sum\limits_{d\mid n}\,2^{\omega(d)}$.
In fact, one can show also that $$\sigma_0\left(n^k\right)=\sum\limits_{d\mid n}\,k^{\omega(d)}$$ for all $k\in\mathbb{N}_0$ (where $0^0:=1$).  One needs to only count the number of the elements of the set
$$S(n,k):=\Big\{(t,d)\in\mathbb{N}\times\mathbb{N}\,\Big|\,t\mid n^k,\,d\mid n,\text{ and }d\text{ is smallest such that }n^k\mid td^k\Big\}\,.$$
The hidden portion below is a remark, almost completely unrelated to the question.

Maybe, we can use this result to extend $\sigma_0$ onto $\mathbb{Q}_{>0}$, by defining $\sigma_0\left(\frac{1}{n}\right):=\sum\limits_{d\mid n}\,(-1)^{\omega(d)}$ and $\sigma_0\left(\frac{u}{v}\right):=\sigma_0(u)\,\sigma_0\left(\frac{1}{v}\right)$ if $\gcd(u,v)=1$.  For example, if $p$ is a prime natural number and $j\in\mathbb{N}_0$, then $\sigma_0\left(\frac{1}{p^j}\right)=1-j$.  This result is consistent with a more natural definition $\sigma_0\left(\prod\limits_{i=1}^\ell\,p_i^{\alpha_i}\right):=\prod\limits_{i=1}^\ell\,\left(1+\alpha_i\right)$ if $p_1,\ldots,p_\ell$ are pairwise distinct prime natural numbers and $\alpha_1,\ldots,\alpha_\ell\in\mathbb{Z}$.  We also have the identity $$\sigma_0\left(n^k\right)=\sum\limits_{d\mid n}\,k^{\omega(d)}$$ for all $n\in\mathbb{N}$ and $k\in\mathbb{Z}$. This identity can be yet again extended in the form $$\sigma_0\left(r^k\right)=\sum\limits_{q\mid r}\,s(q)\,k^{\omega(q)}$$ for all $r\in\mathbb{Q}_{>0}$ and $k\in\mathbb{Z}$ as follows.  For $q,r\in\mathbb{Q}_{>0}$, we say that $q\mid r$ if $q=\frac{u}{v}$ and $r=\frac{x}{y}$, with $u,v,x,y\in\mathbb{N}$, $\gcd(u,v)=1$, and $\gcd(x,y)=1$, satisfy $u\mid x$ and $v\mid y$.  For $q\in\mathbb{Q}_{>0}$, $\omega(q):=\omega(u)+\omega(v)$ and $s(q):=(-1)^{\omega(v)}$ if $q=\frac{u}{v}$ with $u,v\in\mathbb{N}$ and $\gcd(u,v)=1$.  Indeed, if $\sigma_z(n):=\sum\limits_{d\mid n}\,n^z=\prod\limits_{i=1}^\ell\,\left(\frac{p_i^{z\left(1+\alpha_i\right)}-1}{p_i^z-1}\right)$, where $z\in\mathbb{C}\setminus\{0\}$ and $n=\prod\limits_{i=1}^\ell\,p_i^{\alpha_i}$ with distinct primes $p_1,\ldots,p_\ell$ and positive integers $\alpha_1,\ldots,\alpha_\ell$, then we obtain a similar identity $$\sigma_z\left(n^k\right)=\sum\limits_{d\mid n}\,\left(\frac{n}{d}\right)^{kz}\,\varpi_z^k(d)\,,$$ where $\varpi_z^k(n):=\prod\limits_{\substack{{p\mid n}\\p\text{ is prime}}}\,\left(\frac{p^{kz}-1}{p^z-1}\right)$ for $z\neq 0$ and $\varpi_0^k(n):=k^{\omega(n)}$.  We extend $\sigma_z$ onto $\mathbb{Q}_{>0}$ similarly.  Furthermore, if $q=\frac{u}{v}$ where $u,v\in\mathbb{N}$ with $\gcd(u,v)=1$, then we define $\varpi^k_z(q):=\varpi^k_z(u)\,\varpi^k_z(v)$ and $s_z(q):=\prod\limits_{\substack{{p\mid v}\\{p\text{ is prime}}}}\,\left(\frac{-1}{p^z}\right)=\frac{s(q)}{\big(\text{rad}(v)\big)^z}$.  Finally, we get the identity $$\sigma_z\left(r^k\right)=\sum\limits_{q\mid r}\,s_z(q)\,\left(\frac{r}{q}\right)^{kz}\,\varpi^k_z(q)$$ for all rational numbers $r>0$, for every integer $k$, and for every complex number $z\neq 0$.

