Proving $n!$ is never divisible by $2^n$

By Legendre's formula,

$$\nu_2(n!)=\sum_{k=1}^\infty{\left\lfloor\frac n{2^k}\right\rfloor}.$$

But

$$\sum_{k=1}^\infty{\left\lfloor\frac n{2^k}\right\rfloor}<\sum_{k=1}^\infty\frac n{2^k}=n.$$

The strict inequality is justified by the fact that all the terms with $2^k>n$ are missing.

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Note that the "worst case" is when $n$ is a power of $2$: then $n!$ is divisible by $2^{n-1}$.

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Induction:

The development of $n!$ has $\lfloor n/2\rfloor$ even factors. If you discard the odd factors and divide all even ones by $2$, you end-up with the development of $\lfloor n/2\rfloor$!

Hence $$ \nu_2(n!)=\left\lfloor\frac n{2}\right\rfloor+ \nu_2\left(\left\lfloor\frac n{2}\right\rfloor !\right)$$ giving Legendre's formula by induction. (Also using $\lfloor\lfloor n/2^k\rfloor/2\rfloor=\lfloor n/2^{k+1}\rfloor$.)