prime factorization of number which is product of all divisors of another number 
I need to calculate prime factorization of a number which is product of all divisors of a number represented as product of primes.

For example $2^1 \times 3^1 \times 5^1 = 30$, and product of all divisors of 30 is $2 \times 3 \times 5 \times 6 \times 10 \times 15 \times 30 = 2^x \times 3^y \times 5 ^ z$ , I need to calculate $x,y,z$ which are $4,4,4$ in this example 
Is it possible to calculate $x,$y,z without calculating actual product of divisors ? ( may be with divisor function ?)
 A: Let $n=\Pi_i p_i^{e_i}$, where each $p_i$ is a distinct prime factor of $n$ and $e_i$ its exponent. The question is: with which exponent does $p_i$ appear in the product of all divisors of $n$?
Let us first focus on the number $n/(p_i^{e_i})$, and let's count how many distinct divisors it has. Since a generic $p_j\neq p_i$ can appear in each such divisor with exponent $0,\dots,e_j$, there are obviously $\Pi_{j\neq i} (e_j+1)$ distinct divisors of $n/(p_i^{e_i})$. For each such divisor, you have $e_i+1$ distinct divisor of $n$, where $p_i$ appears respectively with exponent $0,...,e_i$. Then, the exponent with which $p_i$ appears in the product of all divisors of $n$ is:
$(\sum_{h=0}^{e_i} h)\Pi_{j\neq i} (e_j+1)=\frac{e_i(e_i+1)}{2}\Pi_{j\neq i} (e_j+1)=\frac{e_i}{2}\Pi_j (e_j+1)$

Example: if $n=1200=2^4\cdot 3^1 \cdot 5^2$, then $(4+1)(1+1)(2+1)=30$ and the product of all divisors of $n$ is $2^{\frac{4}{2}\cdot 30}3^{\frac {1}{2}\cdot 30}5^{\frac{2}{2}\cdot 30}=15407021574586368000000000000000000000000000000$.

This highlights an interesting property of the product of all divisors of $n$ (which has a more elegant and direct proof here). If we denote by $D_n$ the set of all divisors of $n=\Pi_i p_i^{e_i}$, then (see above) its cardinality $|D_n|$ equals $\Pi_j (e_j+1)$, so the product of all divisors of $n$ equals $\Pi_{i} p_i^{\frac{e_i}{2}|D_n|}=n^{\frac{|D_n|}{2}}$.
