How do I prove that there infinitely many rows of Pascal's triangle with only odd numbers?

Modulo $2$,

$$(1+x)^{2^n-1} = \frac{(1+x)^{2^n}}{1+x} = \frac{1+x^{2^n}}{1+x}=1+x+x^2+ \dots + x^{2^n-1}. \qquad\blacksquare$$

Also, modulo an odd prime $p$,

$$(1+x)^{p^n-1} = \frac{(1+x)^{p^n}}{1+x} = \frac{1+x^{p^n}}{1+x}= \frac{1-(-x)^{p^n}}{1-(-x)} = 1 - x+x^2- \dots + x^{p^n-1},$$

which shows that $${p^n-1 \choose k} \equiv (-1)^k \pmod p.$$

illustration purposes; certainly seems rows $2^k - 1$ are a good bet. I would revise the problem and point out that all even, except the initial and final $1,$ seems to be rows $2^k$ only.

I drew this initially for finding the total number of gifts in "The Twelve Days of Christmas" song. I remember telling my father about that twenty or thirty years ago, and him complaining "Why is it binomial (14,3)? It's 12 days, how does 14 make sense?" Much later, I got some graph paper (quadrille) and made the most legible version i could and made a jpeg. I like, especially, how row 14 has consecutive coefficients 1001,2002,3003. Good reason for this, relying on $7 \cdot 11 \cdot 13 = 1001.$

hint

this property follows from the fact that if $m+n=2^k-1$ there are no "carries" in the addition.

the power of $2$ which divides $a!$ is simply $a-\sigma_2(a)$ where $\sigma_2(a)$ returns the sum of the binary digits of $a$

$\binom{2^n-1}{k}=\frac{(2^n-k)\cdot\dots (2^n-2)\cdot(2^n-1)}{1\cdot 2 \cdot 3 \dots k}$ notice that $\bmod 2^n$ the last number on top is congruent to the first multiplied by minus $1\bmod 2^n$, the second to last in the top is congruent to the second in the bottom multiplied by minus one $\bmod 2^n$. Since there is no factor divisible by $2^n$ in the division the congruence $\bmod 2^n$ gives us all the divisibility we need $\bmod 2^n$. Since $m$ and $-m$ are equally divisible by $2^n$ they cancel out as I said.