<s>[INST] Why is row $n = 2^x$ in Pascal's triangle have all even numbers except the $1$'s? In row $n = 2^x$, $x$ being a positive integer, in the Pascal's triangle, all entries except the two $1$'s in extreme left and right are even.
I tried to prove but I couldn't.
Here is my try:-
Every entry is of the form $\frac{(2^x)!}{(k!)([2^x]-k)!}$
I counted the no.of 2's in the prime factorisation of $(2^x)!$ in the following way:-
No.of multiples of $2 = \frac{2^x}2 = 2^{x-1}$
No.of multiples of $4 = \frac{2^x}{2^2}= 2^{x-2}$
Similarily upto no.of multiples of $2^x = 1$.
So total no.of 2's in prime factorisation is
$$
2^{x-1} + 2^{x-2} + \cdots + 2 + 1 = 2^x-1
$$
But I cannot prove that the no.of 2's in the denominator of each entry will be less than $2^{x-1}$.
Can I get some hints/help?
Thank you.
 [/INST] We'll write the greatest integer $\le y$ as $\lfloor y\rfloor$, and define $\{y\}:=y-\lfloor y\rfloor$. The greatest integer $m$ for which $2^m|F!$ is $\sum_{j\ge 0}\left\lfloor\frac{F}{2^j}\right\rfloor$. (This sum's nonzero terms are those with $j\le\lfloor\log_2F\rfloor$.) Thus the greatest integer $m$ for which $2^m|\binom{2^x}{k}$ is $$M:=\sum_{j=0}^x\left(\left\lfloor\frac{2^x}{2^j}\right\rfloor-\left\lfloor\frac{k}{2^j}\right\rfloor-\left\lfloor\frac{2^x-k}{2^j}\right\rfloor\right)=\sum_{j=0}^x\left(\left\{\frac{k}{2^j}\right\}+\left\{\frac{2^x-k}{2^j}\right\}\right).$$This is positive if $1\le k\le2^x-1$.
</s>