Why is row $n = 2^x$ in Pascal's triangle have all even numbers except the $1$'s?

The key observation is that if $1\le k < 2^x$, then $k$ and $2^x-k$ contain exactly the same power of $2$. So when we write

$$\binom{2^x}{k}=\frac{2^x(2^x-1)\ldots(2^x-k+1)}{k(k-1)\ldots2\cdot 1}$$

we can cancel the powers of $2$ in $(2^x-r)$ in the numerator with the powers of $2$ in $r$ in the denominator, leaving us with just the powers of $2$ contained in $$\frac{2^x}{k}$$

which is obviously even if $1\le k < 2^x$.

Here's a simple solution,

$$(1+x)^2 \equiv 1+x^2 \mod 2$$ $$\rightarrow (1+x)^{2^2} \equiv (1+x^2)^2 \equiv 1+x^4 \mod 2$$ if you keep going,

$$(1+x)^{2^n}\equiv 1+x^{2^n}\mod 2$$

implying that all the binomials of the form $\binom{2^n}{k}\equiv 0 \mod 2\qquad \forall 0<k<n$

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For more in-depth analysis I suggest you have a look at Lucas's theorem

The theorem says that for any prime $p$ and numbers $m,n$ which can be written as,

$$m = m_k p^k + m_{k-1}p^{k-1}+ \dots+m_0$$ $$n = n_k p^k + n_{k-1}p^{k-1}+ \dots+n_0$$

we have,

$$\binom{m}{n} = \binom{m_k}{n_k}\binom{m_{k-1}}{n_{k-1}}\dots \binom{m_{0}}{n_{0}} \mod p$$ $2^x$ in binary representation is $10...0$ ($x$ zeroes). Any number $0<k<2^n$ will have binary representation $0b_1b_2...b_x$ where $b_i\in (0,1)$ but not all are 0.

$$\binom{2^x}{k}\equiv \binom{1}{0}\prod_{i=1}^{i=x}\binom{0}{b_i}\mod 2$$

We have $\binom{m}{n}=0$ for $m<n$. Since all the $b_i$ cannot be zero simutaneously ($k\neq 0$ or $2^x$) we have,

$$\binom{2^x}{k}\equiv 0 \mod 2 \qquad 0<k<2^x$$