TITLE: An upper bound for $\log \operatorname{rad}(n!)$
QUESTION [14 upvotes]: Let $n>1$ be an integer and let $\operatorname{rad}(n!)$ denote the radical of $n$-factorial. (The radical of an integer $m$ being, loosely speaking, the product of the prime divisors of $m$.)
Can one give an upper bound for $\log \operatorname{rad}(n!)$ better than $n\log n$? 
Proof: $$ \log \operatorname{rad}(n!) \leq \log (n!) \leq \log (n^n) \leq n \log n. $$

REPLY [6 votes]: A short proof:  Notice that all the primes $p$ with $n<p\leq 2n$ must divide $\binom{2n}{n}$ so that $\frac{\text{rad}((2n)!)}{\text{rad}(n!)}\leq \binom{2n}{n}$.  Also, $\binom{2n}{n}\leq 4^n$.  Consequently for $n=2^k$
$$\text{rad}(n!)\leq \binom{n}{n/2}\cdot \binom{n/2}{n/4}\cdots \binom{2}{1}\leq 4^\frac{n}{2} \cdot 4^\frac{n}{4}\cdots 2\leq 4^n$$ 
and we see $\log \text{rad}(n!)\leq n\log 4$ for $n=2^k$.  You can get other $n$ by noticing $\log \text{rad}(n!)$ is monotonic.  (Without cleverness you could lose a factor of $2$ at worst)