Proving that $2^n$ does not divide $n!$
Idea: Count explicitly how many factors $2$ the numbers in $[1..n]$ contribute to $n!$.

Observe that every other number adds one (the even numbers), every fourth adds another (those divisible by four), every eighth another, and so on.
Hence, the number $\#_2(n!)$ of factors $2$ in $n!$ fulfills

$\qquad\displaystyle\begin{align*}
  \#_2(n!) &amp;\leq \sum_{i=1}^{\log_2(n)} \frac{n}{2^i} \\
           &amp;= n \cdot \sum_{i=1}^{\log_2(n)} \frac{1}{2^i} \\
           &amp;= n \cdot \frac{n-1}{n} \\
           &amp;= n-1 \;.
\end{align*}$

Therefore, $2^n$ can not be a divisor of $n!$.