<s>[INST] Prove that the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than$2^{(a-1)n}$ 
Let $a,n$ be positive integers. Find all $a$ such that for some $n$ the largest power of $2$ dividing $(n+1)(n+2) \cdots (an)$ is greater than $2^{(a-1)n}$.

Since I thought there were no such $a$, I thought about proving this by contradiction. That is, assume that $2^{(a-1)n+1}$ divides $(n+1)(n+2) \cdots (an)$. How can we get to a contradiction from here?
 [/INST] There are $(a-1)n$ factors in the original expression.
$\lfloor \frac {(a-1)n + 1}{2}\rfloor$ are divisible by $2.$
$\lfloor\frac {(a-1)n + 1}{4}\rfloor$ are divisible by $4$.
$\lfloor\frac {(a-1)n + 1}{2^i}\rfloor$ are divisible by $2^i$
The largest power of $2$ that divides $(n+1)\cdots(an) = 2^{\left(\lfloor \frac {(a-1)n + 1}{2}\rfloor + \lfloor \frac {(a-1)n + 1}{2^2}\rfloor \cdots +\lfloor \frac {(a-1)n + 1}{2^i}\rfloor\cdots+\lfloor \frac {(a-1)n + 1}{2^{k}}\rfloor\right)}$
Where k is the largest integer $\le \log_2 ((a-1)n + 1) $
$\sum_\limits{i=1}^{k} \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < ((a-1)n + 1)\sum_\limits {i=1}^{\infty} \frac 1{2^i}\\\sum_\limits {i=1}^{\infty} \frac 1{2^i} = 1\\
\sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor < (a-1)n + 1\\
\sum_\limits{i=1}^{k } \lfloor \frac {(a-1)n + 1}{2^i} \rfloor \le (a-1)n$
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