Prove that all entries of the nth row of Pascal's Triangle are odd given that $n=2^k-1$

Question

Like the title says, I'm to prove that if $n=2^k-1$ then all entries of the nth row of Pascal's Triangle are odd. I know a couple of things that I can use for this, for one I know that I can describe the $j$th entry as $\binom{n}{j}$, and I know that $\binom{n+1}{j} = \binom{n}{j-1} + \binom{n}{j}$. I think the way to go is to show that $\binom{n}{j-1}$ and $\binom{n}{j}$ have opposite parities, thereby showing that their sum must be odd. The clincher is that I'm only allowed to use direct or contrapositive proof, i.e. no induction allowed.

Answer 1

I'd do it this way: the explicit formula for an entry is

$$\binom nj = \dfrac{n!}{j!(n-j)!} = \dfrac{n(n-1)\cdots(n-(j-1))}{j!} = \dfrac{(2^k-1)(2^k-2)\cdots(2^k-j)}{1\cdot 2\cdots j}$$

$2$ divides $2^k-i$ the same number of times it divides $i$ for $1 \leq i \leq j$, so all factors of $2$ on the top and bottom will cancel.