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.

First note that in row 2 of Pascal's triangle is $1,2,1$, so the statement is true for $n=2$. Now let $x$ be arbitrarily chosen and suppose the statement holds for $n=2^{x}$.
In particular this means that for 
$$(1+x)^{2^{x}}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+a_{n}x^{n}$$
we have $a_{1},...,a_{n-1}$ are all even numbers and $a_{0}=a_{n}=1$. So
$$(1+x)^{2^{x+1}}=b_{0}+b_{1}x+b_{2}x^{2}+...+b_{2n-1}x^{2n-1}+b_{2n}x^{2n}$$
$$=(1+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+x^{n})^{2}$$
$$=1+x^{2n}+\sum^{n-1}_{i=1}a_{i}^{2}x^{2i}+\sum_{0\leq i<j\leq n}2a_{i}a_{j}x^{i+j}$$
Since for $0\leq i<j\leq n$ we find that $0<i+j<2n$ we find that $b_{0}=b_{2n}=1$ and for $0<k<2n$ we find that $b_{k}$ is the sum of even numbers and therefore even.
