Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example,
$$p(5184)=p(2^{6}\cdot 3^{4})=24,\qquad
p(65536)=p(2^{16})=16.$$
Is $p(n) = O(\log^{k}(n))$ for some constant $k$? Thanks in advance!



Upper bound for product of exponents of prime factorization
No, $p(n)$ is not bounded by a power of $\log(n)$. For example, if $$n=p\_1^2\dots p\_r^2,$$ 
where $p\_1<p\_2<\dots$ is the sequence of prime numbers, then by the Prime Number Theorem we have
$$ \log n=2(\log p\_1+\dots +\log p\_r)=(2+o(1))p\_r=(2+o(1))r\log r, $$
so that $r=(1/2+o(1))\log n/\log\log n$, and therefore
$$ p(n)=2^r = \exp\left(\frac{(\log2+o(1))(\log n)}{2\log\log n}\right).$$
Note that the exponent on the right hand side grows faster than any constant times $\log\log n$.


In fact the right hand side is a general upper bound for $p(n)$. To see this, consider a general $n$ with prime power decomposition 
$$n=q\_1^{k\_1}\dots q\_r^{k\_r},$$ 
where each $q\_j$ is a prime and each $k\_j$ is a positive integer. Let $m>0$ be a parameter, then
$$ p(n)=\prod\_{1\leq j\leq r} k\_j\leq 
\prod\_{\substack{1\leq j\leq r\\q\_j\leq m}}k\_j
\prod\_{\substack{1\leq j\leq r\\q\_j>m}}2^{k\_j/2}\leq 
\left(\frac{\log n}{\log 2}\right)^m 
\left(\prod\_{\substack{1\leq j\leq r\\q\_j>m}} q\_j^{k\_j/2}\right)^{\frac{\log 2}{\log m}},$$
whence
$$ p(n)\leq(\log n)^m n^\frac{\log 2}{2\log m}=\exp\left(m\log\log n+\frac{(\log 2)(\log n)}{2\log m}\right).$$
Choosing $m:=\log n/(\log\log n)^3$ yields that
$$ p(n)\leq \exp\left(\frac{(\log2+o(1))(\log n)}{2\log\log n}\right). $$
This argument was inspired by Section I.5.2 of Tenenbaum: Introduction to analytic and probabilistic number theory (Cambridge University Press, 1995).


