<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] Suppose we are looking at the binomial coefficient $$\binom {2^n}m=\frac {(2^n)!}{m!(2^n-m)!}=\frac {2^n}m\cdot\frac {2^n-1}1\cdot\frac{2^n-2}2\dots \cdot \frac {2^n-(m-1)}{m-1}$$
(assuming $m\gt 1$: we have only the first term for $m=1$, and for $m=0$ the empty product or straightforwardly the value $1$)
Now suppose $2^n\gt r \gt 0$ and that $r=2^st$ where $t$ is odd, and we have $s\lt n$, then all the fractions in the product apart from the first are of the form $$\frac {2^n-r}r=\frac {2^n-2^st}{2^st}=\frac {2^{n-s}-t}{t}$$and this is a fraction with odd numerator and denominator. On the other hand with $\frac {2^n}m$ we have that $m$ is divisible by a lower power of $2$ than $2^n$ unless $m=2^n$. So we have a positive power of $2$ in the product.
This can be adapted for powers of any prime.
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