Prove that $2^n$ does not divide $n!$

Question

I want to prove that $2^n$ does not divide $n!$.

I was trying by induction and I'm confused about if what I'm doing is right.

First I test it with $n=1$. In fact: $$2^1 \nmid 1!$$

So if i take the I.H. as $2^n \nmid n!$ and I try to prove it for $n+1$: $$2^{n+1} \nmid (n+1)!$$ $$2^{n} \cdot 2 \nmid (n+1) \cdot n!$$

As $2^n \nmid n!$ it must be that $2^n$ divides $n+1$, so I need to prove that it doesn't. If I try by induction again I must have $n>1$ for it to be true:

$$P(n) = 2^n \nmid n+1$$ $$P(n+1) = 2^n \cdot 2 \nmid n+1+1$$ but $2^n \nmid n+1$ and $2^n \nmid 1$ because $n>1$

So my question is obviously if this is correct. I'm doubting because the exception I have to do with $n$ being greater than 1 for the second part. If I made a mistake could you point to me in a better direction? I'm sure there must be a simpler way to prove this.

Thanks!

Answer 1

The maximal power of $2$ which divides $n!$ is $$v_2(n!)=\lfloor \frac n2 \rfloor+\lfloor \frac n4 \rfloor +\cdots=\sum_{i=1}^{\infty} \lfloor\frac n{2^i}\rfloor$$

We bound this from above by dropping the floor function to see that $$v_2(n!)<n\sum_{i=1}^{\infty} \frac 1{2^i}=n$$

Answer 2

By Legendre's theorem $$\nu_2(n!) = \sum_{m\geq 1}\left\lfloor\frac{n}{2^m}\right\rfloor\leq \sum_{m\geq 1}\frac{n}{2^m}=n. $$ Equality may hold only if $2^k\parallel n$, but in such a case the $(k+1)$-th terms of the central sums fullfill a strict inequality.

Answer 3

A proof by induction will be difficult because, as $n$ increments, $2^n$ adds one factor of $2$ while $n!$ can add many.

The way I would do it is use Legendre's theorem, stated here, for example: http://www.cut-the-knot.org/blue/LegendresTheorem.shtml

If $p$ is a prime, and $v_p(n!)$ is the exponent of the greatest power of $p$ dividing $n!$, so $p^{v_p(n!)} \mid n!$ and $p^{v_p(n!)+1} \not\mid n!$, then $v_p(n!) =\sum_{k=1}^{\lfloor \log_p n \rfloor} \lfloor \dfrac{n}{p^k} \rfloor $.

It follows that

$\begin{array}\\ v_p(n!) &=\sum_{k=1}^{\lfloor \log_p n \rfloor} \lfloor \dfrac{n}{p^k} \rfloor\\ &\le\sum_{k=1}^{\lfloor \log_p n \rfloor} \dfrac{n}{p^k} \\ &=n\sum_{k=1}^{\lfloor \log_p n \rfloor} \dfrac{1}{p^k} \\ &<n\sum_{k=1}^{\infty} \dfrac{1}{p^k} \qquad\text{(here is where we get a strict inequality)}\\ &=n\dfrac{\frac1{p}}{1-\frac1{p}}\\ &=\dfrac{n}{p-1}\\ \end{array} $

In particular, for $p=2$, $v_2(n!) < n $, so $2^n \not\mid n! $.

Note that if $n=p^m$ for some integer $m$, then

$\begin{array}\\ v_p(n!) &=v_p((p^m)!)\\ &=n\sum_{k=1}^{\lfloor \log_p n \rfloor} \dfrac{1}{p^k}\\ &=n\sum_{k=1}^{m} \dfrac{1}{p^k}\\ &=p^m\dfrac{\frac1{p}-\frac1{p^{m+1}}}{1-\frac1{p}}\\ &=p^m\dfrac{1-\frac1{p^{m}}}{p-1}\\ &=\dfrac{p^m-1}{p-1}\\ \end{array} $

If $p=2$, $v_2((2^m)!) =2^m-1 $, so this misses by just $1$, so that $2^{2^m-1} \mid (2^m)!$ and $2^{2^m} \not\mid (2^m)!$.

Answer 4

The mistake in your work is the conclusion that if $2^n | j * k$, then either $2^n | j$ or $2^n | k$. For instance, consider that 8 divides $12*6 = 72$ but neither 12 or 6 is divisible by 8.

Answer 5

As pointed out in the other answers we can write the maximum power of $2$ that divides $n!$ as $\nu_2(n!) = \sum_{k\geq 0}\left\lfloor\frac{n}{2^k}\right\rfloor$. We can give a simple interpretation of this number by working in binary. The binary expansion of $n$ is $n = \sum_{k\geq 0} a_k 2^k$ with $a_k\in\{0,1\}$. Using this we find

$$\matrix{\nu_2(n!) &=& \sum_{k\geq 0}\left\lfloor\sum_{i\geq 0} a_i 2^{i-k}\right\rfloor \\= \sum_{k\geq 0}\sum_{i\geq k} a_i 2^{i-k} &=& \sum_{k\geq 0} a_k(1+2+4+\ldots + 2^{k-1}) \\= \sum_{k\geq 0} a_k(2^k-1) &=& n - \sum_{k\geq 0} a_k}$$

Thus

$$\nu_2(n!) = n - s_2(n)$$

where $s_2(n)$ is the sum of the binary digits of $n$. If $n>0$ then the binary expansion of $n$ must contain atleast one "$1$" so $\nu_2(n!) < n$ follows. The minimum value of $s_2(n)$ for $n>0$ occurs if $n$ is on the form $2^k$ for which $s_2(n) = 1$ and $v_2(n!) = n-1$.