When does $2^{(k-1)n+1}$ not divide $\dfrac{(kn)!}{n!}$? 
Find all positive integers $k$ such that for any positive integer $n$, $2^{(k-1)n+1}$ does not divide $\dfrac{(kn)!}{n!}$.

We have $$\dfrac{(kn)!}{2^{(k-1)n+1} \cdot n!} = \dfrac{(n+1)\cdots(kn)}{2^{(k-1)n+1}}.$$ Thus we need the largest power of $2$ dividing $(n+1)\cdots(kn)$ for some $k$ to be less than $2^{(k-1)n+1}$ for all $n$. How can we find all such $k$?
 A: Only the powers of $2$ work:
There is a theorem that says that $v_p(n!)=\frac{n-S_p(n)}{p-1}$, where $S_p(n)$ is the sum of the digits in the base $p$ representation of $n$. Using this we get $v_2\frac{(kn)!}{n!}=(k-1)n+(S_p(n)-S_p(kn) )$.
So  the number $k$ must satisfy that $kn$ has sum of digits greater than or equal to $n$ for any $n$ (in binary).
Notice that $k$ satisfies this if and only if $2k$ satisfies this, so we must only prove that the only odd integer that works is $k=1$.
Suppose an odd $k>1$ works. Let $m$ be a multiple of $k$ with the least possible number of one digits (suppose it has $v$ ones), we can clearly take $m$ to be odd. Write $m=2^a+b$ with $2^a>b$. Notice that the number $w=2^{a+s\varphi(k)}+b$ is also a multiple of $k$ with $v$ ones for every $s\in \mathbb Z^+$.
We now prove that $w/k$ must have more than $v$ ones for sufficiently large $s$.We proceed by contradiction, pick $s$ so that $w/k$ 
must have at least $z$ consecutive zeros in its binary representation , where $z$ is equal to the length of the base $2$ representation of $k$ (it must for large values of $w$, because we are assuming $w/k$ has at most $v$ one digits, and $w/k$ is odd). Then the number of one digits in $w/k\times k$ is equal to the sum of the bits of $k$ multiplied by each of the two parts of $w/k$, and this is at least $2v$ (here we consider the two parts to be the digits corresponding to each of the two sides separated by the $z$ consecutive zeros). This is a contradiction.
