The number of positive integers $n$ such  that $2^n$ di vides n! is 

The exponent of 2 in $n!$ is given by

$E=\left[\frac{n}{2}\right]+\left[\frac{n}{2^2}\right]+\left[\frac{n}{2^3}\right]+\dots$

where [x] denotes greatest integer  $\le x,\text{ As }\left[x\right]\le x\mathrm{\forall }x$

$\left[\frac{n}{2^m}\right]=0$ after finite number of terms. Thus we get

$E<\frac{n}{2}+\frac{n}{2^2}+\frac{n}{2^3}+\dots =\frac{n/2}{1-1/2}=n$

Thus, there is no positive integer for which $2^n$ divides $n!$