TITLE: Prove: $\forall$ $n\in \mathbb N, (2^n)!$ is divisible by $2^{(2^n)-1}$ and is not divisible by $2^{2^n}$
QUESTION [6 upvotes]: I assume induction must be used, but I'm having trouble thinking on how to use it when dealing with divisibility when there's no clear, useful way of factorizing the numbers.

REPLY [2 votes]: A Combinatorial Approach:
For each nonnegative integer $n$, show that $a_n:=\frac{\left(2^n\right)!}{2^{2^n-1}}$ is the number of ways to label the leaves of a complete binary tree $T_n$ with $2^n$ leaves by $1$, $2$, $\ldots$, $2^n$.  (Two labelings are considered to be the same if there is a graph automorphism $f$ on $T_n$ such that, for $i=1,2,\ldots,2^n$, $f$ sends the leaf with label $i$ in one labeling to the leaf with the same label in the other labeling.)  Prove that $a_n$ is odd for all $n\in\mathbb{N}_0$ by showing that $\frac{a_n}{a_{n-1}}$ is an odd integer for $n=1,2,3,\ldots$, whereas $a_0=1$.  (This second part can be proven combinatorially as well, because $\frac{a_n}{a_{n-1}}$ for $n\in\mathbb{N}$ is the number of ways to partition $\left\{1,2,\ldots,2^n\right\}$ into $2^{n-1}$ subsets each of which has $2$ elements.  It follows immediately that $\frac{a_n}{a_{n-1}}=\left(2^n-1\right)!!=\prod_{i=1}^{2^{n-1}}\,(2i-1)$.)