Show that $\dbinom{2^{n}-1}{2^{n-1}}$ is an odd number I would appreciate if somebody could help me with the following problem

Show that
$$\dbinom{2^{n}-1}{2^{n-1}}$$
is an odd number, where $n$ is a positive integer.

I tried to solve this problem but I can't.
 A: Here is a trick.
You are looking at $$\binom {2^n-1}{2^{n-1}}=\frac {(2^n-1)!}{2^{n-1}!(2^{n-1}-1)!}=\frac {(2^n-1)(2^n-2) \dots (2^{n-1}+1)}{(2^{n-1}-1)\dots 3\cdot 2\cdot 1}$$
Now look at the factors in the numerator - say $a=2^n-r$ and consider the highest power of $2$ - say $2^t$ which is a factor of $a$. We have $t\lt n$ whence $2^t\mid 2^n-a=r$, (and incidentally it is easy to see that $2^{t+1} \nmid 2^n-a$).
If we are concerned with divisibility by powers of $2$ only, we can therefore replace each factor $2^n-r$ in the numerator by $r$ (and without the incidentally part, note that the power of $2$ involved cannot decrease), and you will see that the expression reduces trivially to $1$. Since the original number was an integer and the parity has not been changed, the original number must have been odd (and without the incidentally comment, we note that the the power of $2$ did not increase, so must have been zero all along, and the correspondence of powers must have been exact).
A: Every binomial coefficent $\dbinom{n}{k}$
  with $0\leq k \leq n$
  is not divisible by $p$
  iff $n+1$ is divisible by $p^{d}$
  for some $d$ and so your binomial coefficient is not divided by $2$. You can find a proof here. 
A: $$\binom{2^n-1}{2^{n-1}}=\frac{(2^n-1)\cdot(2^n-2)\cdot\ldots\cdot 2^{n-1}}{2^{n-1}!}=\frac{(2^n-1)!}{2^{n-1}!(2^{n-1}-1)!}$$
and:
$$ \nu_2(2^{n-1}!) = 2^{n-2}+2^{n-3}+\ldots+1 = 2^{n-1}-1, $$
$$ \nu_2((2^{n-1}-1)!) = 2^{n-1}-n, $$
$$ \nu_2((2^n-1)!) = 2^n-n-1,$$
hence:

$$\nu_2\binom{2^n-1}{2^{n-1}}=0$$

as wanted. Here we took $\nu_2(m)=\max\left\{h\in\mathbb{N}:2^h\mid m\right\}$ and exploited:
$$ \nu_p(n!) = \left\lfloor\frac{n}{p}\right\rceil+\left\lfloor\frac{n}{p^2}\right\rceil+\left\lfloor\frac{n}{p^3}\right\rceil+\ldots $$
A: More in general $\binom{2^n-1}{k}$ is odd for every $k\leq2^n-1$ infact:
We can write
$$\binom{2^n-1}{k}=\frac{(2^n-1)(2^n-2)......(2^n-k)}{k!}$$Now note this: For every $i\in\{1.....k\}$ the factorization of $2^n-i$ contains the same powers of 2 that contains $i$ infact suppose $i=2^tc$ where c is odd then $2^n-i=2^n-2^tc=2^t(2^{n-t}-c)$.
i.e  if $$2^t||i\Rightarrow 2^t||(2^n-i)$$
Then simplifying the fraction from powers of 2 we obtain an odd number for every k.
A: You must compute the $2$-valuations of $(2^n-1)!$, $\,(2^{n-1})!$ and $\,(2^{n-1}-1)!$. For  this, you have Legendre's formula: for any prime $p$,
$$ v_p\big((n)!\bigr)=\sum_{k\ge 1}\biggl\lfloor\frac{n}{p^k}\biggr\rfloor.$$
If $n=p^r$, it simplifies to:
$$v_p\bigl((p^r)!\bigr)=p^{r-1}+p^{r-2}+\dots+1=\frac{p^r-1}{p-1}$$
Finally, if $p=2$, we have:
$$v_2\bigl((2^r)!\bigr)=2^r-1, \enspace\text{hence:}\quad v_2\bigl((2^r-1)!\bigr)=2^r-r-1.$$
Now we have all ingredients for the computation:
$$ v_2\biggl(\binom{2^n-1}{2^{n-1}}\biggr)= 2^n-(n+1) - (2^{n-1}-1)-(2^{n-1}-n)=0.$$
Thus $\dbinom{2^n-1}{2^{n-1}}$ is not divisible by $2$.
A: $\dbinom{2^n-1}{2^{n-1}}=\dfrac{(2^n-1)!}{(2^n-1-2^{n-1})!(2^{n-1})!}=\dfrac{(2^n-1)!}{(2^{n-1}-1)!(2^{n-1})!}$.
Expanding the factorials means we can remove all the even terms from the numerator, as $2^n-2j=2(2^{n-1}-j)$, with $1\le j \lt 2^{n-1}$. $\sum j=2^{n-1}-1$, as does $\nu_2(2^{n-1}!)=2^{n-1}-1$
