What does the product of all proper divisors equal to? I'm looking for a formula to calculate the product of all proper divisors of a number number $n$. Any idea?    
Thanks,
Chan
 A: Like Sivaram points out. If $n$ is not a perfect square, then every divisor $d_1$ can be paired with the divisor $\frac{n}{d_1}$, which is distinct from $d_1$; the product of these two is $n$. So the product of all divisors is equal to $n^k$, where $2k$ is the number of divisors. (Note that a positive integer has an odd number of distinct divisors if and only if it is a square).
If $n$ is a perfect square, then it will have $2k+1$ divisors for some $k$; the product will be $n^k\times\sqrt{n} = n^{k+\frac{1}{2}}$. 
Either way, the answer is $n^{d(n)/2}$, where $d(n)$ is the number of divisors of $n$ (sometimes written $\sigma_0(n)$ or $\tau(n)$).
If you want the proper divisors, then you are excluding $n$ from the product, so the answer becomes $n^{(d(n)/2) - 1}$.  
A: We are meant to calculate,
$$
\prod_{t|n}t, t\neq n
$$
We will calculate including $t=n$, and will divide the result finally by  $n$.
Using fundamental theorem of arithmetic,
$$
n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}
$$
where $p_j$ is a prime number. Now if $p^a$ is an entry in the above, number of terms in above product containing $p^{a'}$ is $d(n)/(a+1)$, where $d(n)$ denotes the number of positive divisors of number $n$. Thus in above product, contribution out of prime $p$ is 
$$
p^{(1+2+\cdots  +a)*d(n)/(a+1)}=(p^a)^{d(n)/2}
$$
So, 
$$
\prod_{t|n}t=(p_1^{a_1})^{d(n)/2}(p_2^{a_2})^{d(n)/2}\cdots (p_k^{a_k})^{d(n)/2}=n^{d(n)/2}
$$
Division by $n$ gives,
$$
n^{\frac{d(n)}{2}-1}
$$
