nicetry007 wrote: When $ n \;=\; 2^{s}$, the highest power of $ 2$ that divides $ n!$ is $ \frac{n}{2}\;+\;\frac{n}{4}\;+\;\cdots+\;\frac{n}{2^{s}}\; =\; 2^{s-1}\;+\; 2^{s-2}\;+\;\cdots\;+\;1 \;=\; 2^{s}-1\; = \;n-1$. When $ 2^{s}\; <\; n \;<\; 2^{s+1}$, the highest power of $ 2$ that divides $ n!$ is $ \lfloor\frac{n}{2}\rfloor\;+\;\lfloor\frac{n}{4}\rfloor\;+\;\cdots+\;\lfloor\frac{n}{2^{s}}\rfloor\; \leq \; \frac{n}{2}\;+\;\frac{n}{4}\;+\;\cdots+\;\frac{n}{2^{s}}\; = \;\frac{n}{2}\;\left( 1 \;+\;\frac{1}{2}\;+\;\cdots\;+\; \frac{1}{2^{s-1}}\right)\;$ $ = \; \frac{n}{2}\;\left(\;\frac{1-\frac{1}{2^{s}}}{1-\frac{1}{2}}\;\right) =\; n-\frac{n}{2^{s}}\; < \;n-1$. Hence, $ 2^{n-1}$ divides $ n!$ if and only if $ n$ is a power of $ 2$.

Here is an use full result : ${ v_{p}(n!) = \frac {n - S_{p}(n)}{p - 1}}$ where $ S_{p}(n)$ is sum of digit of n when represent in base p Proof Let $ n = \sum_{i = 1}^k a_ip^i$ where $ 0\leq a_i\leq p - 1$ $ v_{p}(n!) = \sum_{i = 1}^{ + \infty}[\frac {n}{p^i}] = \frac {n - {\sum_{i = 1}^k a_i}}{p - 1}$ This law has a name but I don't remember. Apply this lemma for $ p = 2$ we has : $ v_2(n!) = n - S_{2}(n)$ To $ 2^{n - 1}|n!$ then $ n - 1\leq n - S_{2}(n)$ imply that $ S_2(n)\leq 1$ so $ S_2(n) = 1$ imply that $ n = 2^k$ where $ k\in N$

Peter wrote: Find all $n \in \mathbb{N}$ such that $ 2^{n-1}$ divides $n!$. By Kummer's thereom,we need $\frac{n-S_(n)}{n-1}$ which happens only for $n=2^k$

$n=2^d$.