How to show that $\prod_{d/n} d = n^{\frac{\tau(n)}{2}}$ set $ n, n \in \mathbb{N}$ and prove that 
$\prod_{d/n} d = n^{\frac{\tau(n)}{2}}$
¨I have tried this¨
If $n > 1$ then
$n = p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}$
so 
$n^{\frac{\tau(n)}{2}}=(p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}})^{\frac{(\alpha_{1}+1)\cdot(\alpha_{2}+1)\cdots(\alpha_{k}+1)}{2}}$
but i dont know how stablish a relation with $\prod_{d/n} d$
 A: Square the equation. On the right you get $n^{\tau(n)}$. Now try to pair the factors on the left in a good way.
A: Assume that $n$ is not a square, hence $\tau(n)$ is even. You have:
$$\prod_{d|n} d =\Big (\prod_{\substack{d\leq \sqrt{n}\\ d|n}} d\Big)\cdot \Big (\prod_{\substack{d\geq \sqrt{n}\\ d|n}} d\Big)=\prod_{\substack{d\leq \sqrt{n}\\d|n}}d\cdot\frac{n}{d}=n^{\frac{\tau(n)}{2}},\tag{1}$$
just by "coupling" any divisor $d$ with its complementary divisor $\frac{n}{d}$. With a minor fix, the above argument also works if $n$ is a square ($\sqrt{n}$ is accounted only once in the LHS).
A: If you want to go on in the direction in which you started : 
$\prod_{d/n} d=\prod_{\{(\beta_1, ... , \beta_k) \ , \ 0 \leq(\beta_1,...,\beta_k) \leq (\alpha_1, ...,\alpha_k )\}}(p_{1}^{\beta{1}}\cdot p_{2}^{\beta{2}}\cdots p_{k}^{\beta{k}}) =(\prod_{\beta_1=0}^{\alpha_1} p_1^{\beta_1})^{(\alpha_{2}+1)\cdots(\alpha_{k}+1)}\ \cdot \ ... \ \cdot (\prod_{\beta_k=0}^{\alpha_k} )^{(\alpha_{1}+1)\cdot(\alpha_{2}+1)\cdots(\alpha_{k-1}+1)}p_k^{\beta_k}=(p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}})^{\frac{(\alpha_{1}+1)\cdot(\alpha_{2}+1)\cdots(\alpha_{k}+1)}{2}}$
A: Thanks to all for the help finally this is what i did to answer the problem:
if   $p=\prod_{d/n}d$ 
then 
$p = \prod_{d/n}(\frac{n}{d})$
if we multiply side by side... we have
$p^{2} = \prod_{d/n} d\cdot(\frac{n}{d})= \prod_{d/n} n = n^{\tau(n)}$
so 
$p^{2} =n^{\tau(n)}$ 
and 
$p= n^{\frac{\tau(n)}{2}}$
then 
$\prod_{d/n }d= n^{\frac{\tau(n)}{2}}$ and is done.....
