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 $n =2^k-1$.

Counting the Number of Factors of $\boldsymbol{2}$
In Corollary $(7)$ of this answer, it is shown that the number of factors of $p$ in $\binom{n}{k}$ is
$$
\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ us the sum of the digits in the base-$p$ representation of $n$.
In base-$2$, each digit of $n=2^k-1$ is a $1$. For $0\le k\le n$, each digit of $k$ that is a $0$ is a $1$ in $n-k$ and vice-versa. Thus, $\sigma_p(k)+\sigma_p(n-k)=\sigma_p(n)$. Thus, formula $(1)$ says that there are $0$ factors of $2$ in $\binom{n}{k}$.
That is, $\binom{n}{k}$ is odd.

Induction
We will show that for $n\ge1$,
$$
\binom{2^n}{k}\text{ is even for }0\lt k\lt2^n\tag{2}
$$
and of course $\binom{2^n}{0}=\binom{2^n}{2^n}=1$.
Note that $(2)$ is true for $n=1$. Suppose that $(2)$ is true for some $n$. Then we can use Vandermonde's Identity to get
$$
\binom{2^{n+1}}{k}=\sum_{j=0}^k\binom{2^n}{j}\binom{2^n}{k-j}\tag{3}
$$
where inductively, $\binom{2^n}{j}\binom{2^n}{k-j}$ is even except when either $j=0$ or $j=2^n$ and either $k-j=0$ or $k-j=2^n$. That gives $4$ cases:


*

*$j=k=0$: makes $\binom{2^{n+1}}{0}$ odd

*$j=0$ and $k=2^n$: paired with case 3, makes $\binom{2^{n+1}}{2^n}$ even

*$j=2^n$ and $k=2^n$: paired with case 2, makes $\binom{2^{n+1}}{2^n}$ even

*$j=2^n$ and $k=2^{n+1}$: makes $\binom{2^{n+1}}{2^{n+1}}$ odd
Thus, accounting for these $4$ special cases, $(3)$ shows that $(2)$ holds for $n+1$.

Since $\binom{2^n-1}{0}=1$ and for $0\lt k\lt2^n$
$$
\binom{2^n-1}{k}=\binom{2^n}{k}-\binom{2^n-1}{k-1}\tag{4}
$$
we get that $\binom{2^n-1}{k}$ is odd (even minus odd) for $0\le k\le2^n-1$.
