How can prove that $2^n \nmid n!$using binary representation for $n!$ and $2^n$.
Proving that $2^n$ does not divide $n!$
discrete mathematics;number theory
I was working in counting how many trailing zeros has $n!$ in base-2, it'll be $d$.Because, trailing zeros means that you can divide $n!$ by 2, $d$ times.$$ n! = \underbrace{****1\overbrace{0000\cdots 000}^{d\text{ zeros}}}_{\lfloor \log n!\rfloor + 1 \ \text{ digits}}\ \  \Rightarrow \, 2^{d}\, \vert\, n!\ \ \ \ \text{ with *}\in \{0,1\}$$So, If $T(n)$ counts trailing zeros for $n!$.$$ T(0) = 1\ \ \ T(1) = 0 \ \ \ T(n) = \lfloor n/2 \rfloor + T(\lfloor n/2 \rfloor)$$Or it could be $$ T(n) = \sum _{i=1}^{\log n} \left \lfloor \frac{n }{2^i} \right \rfloor$$ So, there is a counterexample, consider it, $n=2^k$$$ \begin{align}T(n) &=  \sum _{i=1}^{\log n} \left \lfloor \frac{2^k}{2^i} \right \rfloor =\sum _{i=1}^{k}  2^{k-i}  = \sum _{i=1}^{k}  2^{k-i} = 2^k \left ( \sum _{i=1}^{k}  2^{-i} \right )\\& = 2^k \left( \frac{(1/2)-(1/2)^{k+1}}{1 - (1/2)} \right) = \frac{2^{k-1} - 1/2}{1/2} = 2^k -1\end{align}$$then $2^{k} - 1\ \vert\ n!$, and for that $2^{n} \not\vert\ n!$.