<s>[INST] If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd. Prove: If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
I know that the $n$th row in pascal's triangle correspond to the coefficients of $(x+y)^n$:
$$\begin{align}(x+y)^n = {n \choose 0} x^n + {n \choose 1} x^{n-1} y + {n \choose 2}x^{n-2}y^2+{n \choose 3} x^{n-3} + \cdots + {n \choose n-1} xy^{n-1}+{n \choose n} y^n \end{align}.$$ 
I have to show that each coefficient $\begin{align} {n \choose 0},{n \choose 1},{n \choose 2},\ldots,{n \choose n-1},{n \choose n} \end{align}$ is of the form $2b+1$, but I'm not sure how to do this with $n =2^k-1$.
 [/INST] As the answer above says, $\displaystyle{2^{n}-1\choose k-1}+{2^{n}-1\choose k} = {2^n \choose k}$. So if  ${2^n \choose k}$ is even for $k \neq 0,2^n$, both terms are even or both terms are odd. First, let's prove that ${2^n \choose k}$ is even.
$$\displaystyle{2^n \choose k} = \frac{2^n!}{k!(2^n-k)!}$$
For $k = 1, {2^n \choose 1} = 2^n$ even. Now assume ${2^n \choose k}$ is even for some k, now: 
$$\displaystyle{2^n \choose k} = \frac{2^n!}{k!(2^n-k)!} = \frac{2^n!(k+1)}{(k+1)!(2^n-k-1)!(2^n-k)} = {2^n \choose k+1}\frac{k+1}{2^n-k}$$
Now, if $k$ is either odd or even, $\displaystyle\frac{k+1}{2^n-k}$ is odd. As $\displaystyle{2^n \choose k}$ is even, it has to happen that $\displaystyle{2^n \choose k+1}$ must be even. Therefore by induction, $\displaystyle{2^n \choose k}$ is even for all $k \neq 0,2^n$.
Now, by induction as $\displaystyle{2^{n}-1\choose k-1}+{2^{n}-1\choose k} = {2^n \choose k}$, note that, as the first two terms ($k = 0,1$) are odd, then the third term must also be odd, and so on.
Therefore ${2^n-1 \choose k}$ is odd for all $k$ between $0$ and $2^n-1$.
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