TITLE: How do I prove that there infinitely many rows of Pascal's triangle with only odd numbers?
QUESTION [15 upvotes]: This is exercise number $59$ from Chapter $2$ of Hugh Gordon's Discrete Probability. 

Show that there are infinitely many rows of Pascal's Triangle that consist entirely of odd numbers.

Intuitively, if you draw boxes around the numbers in Pascal's triangle and color the boxes black if the number is odd and white if the number is even, then the triangle will look like the Sierpinsky triangle as you zoom out. 
In particular, if we number the rows starting with the top as $1$, the rows will all odd numbers will be exactly the rows with number $2^n$ for some $n \in \mathbb{N}$ (or $n=0$ for the first). You can see this if you think about the Sierpinsky triangle coloring.
Anyway, is there any direct way to show the following for all $k$ with $0 \leq k \leq 2^n-1$?
$$\binom{2^n-1}{k} \equiv 1 \pmod{2}$$
This can probably be done by induction but a direct proof would be preferable.

REPLY [3 votes]: How about a nice combinatorial proof after all that algebra? (It is still by induction, though -- sorry.)
$\binom{2^n-1}{k}$ is the number of strings of $2^n-1$ letters of which $k$ are a and the rest are b.
Most of these strings can be paired with a different string of the same kind by reading it backwards. The ones that can't are the palindromes.
How many palindromes are there? We can choose a palindrome by taking one of the $\binom{2^{n-1}-1}{\lfloor k/2\rfloor}$ possible first almost-halves, followed by either a or b according to whether $k$ is odd or even, followed by the reverse of the first half.
By the induction hypotheis $\binom{2^{n-1}-1}{\lfloor k/2\rfloor}$ is odd. So there's an odd number of palindromes among our strings. And the non-palindromes match up two by two, so adding those in we still have an odd number of qualifying strings.