Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Then the non-zero elements of the $n$th row of Pascal's triangle are all odd :
 * $n = 2^m - 1$

for some $m \in \Z_{\ge 0}$.

As can be seen, the entries in rows $0, 1, 3, 7$ are all odd.

Proof
The statement:
 * $\dbinom n k$ is odd

is equivalent to:
 * $\dbinom n k \equiv 1 \pmod p$

The corollary to Lucas' Theorem gives:
 * $\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$

where:
 * $n, k \in \Z_{\ge 0}$ and $p$ is prime
 * the representations of $n$ and $k$ to the base $p$ are given by:
 * $n = a_r p^r + \cdots + a_1 p + a_0$
 * $k = b_r p^r + \cdots + b_1 p + b_0$

When $p = 2$, the digits $a_j, b_j$ are either $0$ or $1$.

In order for $\dbinom n k \equiv 1 \pmod p$, it is necessary and sufficient that $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $j \in \set {0, 1, 2, \ldots, r}$.

In order for this to happen, either:
 * $\dbinom {a_j} {b_j} = \dbinom 0 0$

or:
 * $\dbinom {a_j} {b_j} = \dbinom 1 0$

or:
 * $\dbinom {a_j} {b_j} = \dbinom 1 1$

Suppose $a_i = 0$ for some $i \in \set {0, 1, 2, \ldots, r}$.

Then if $b_i = 1$:
 * $\dbinom {a_j} {b_j} = \dbinom 0 1 = 0$

and so:
 * $\dbinom n k \equiv 0 \pmod p$

for whichever $k$ (and there will be at least one) has digit $b_i = 1$.

So the only way it can be assured that all $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $k \in \set {0, 1, 2, \ldots, n}$ is for $a_j = 0$ for all $j \in \set {0, 1, 2, \ldots, r}$.

That is, for $n = 2^m - 1$ for some $m \in \Z_{\ge 0}$.