TITLE: Prove that $\frac{(2n)!}{2^n} $ is an integer for all $n \geq 0$
QUESTION [5 upvotes]: I don't really know where to begin with this problem. Should I use induction to solve it or is there an easier way?

REPLY [5 votes]: Consider the number of ways to arrange the numbers $1,1,2,2,\ldots,n,n$ in a line. Standard combinatorics tells you that it is $\frac{(2n)!}{2^n}$, hence that expression is an integer.
More details for why the number of ways is the desired expression: There are $(2n)!$ ways of arranging the list. Since for every $k$ we can switch the two copies of $k$ and be left with the same list, we have overcounted, so we should divide by $2$ for every $k$ to get the correct count. This gives the $2^n$ in the denominator.