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
Lucas' Theorem gives:
 * $\dbinom n k \equiv \dbinom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {n \bmod p} {k \bmod p} \pmod p$

where $n, k \in \Z_{\ge 0}$ and $p$ is prime.

When $p = 2$ this gives:
 * $\dbinom n k \equiv \dbinom {\left \lfloor {n / 2} \right \rfloor} {\left \lfloor {k / 2} \right \rfloor} \dbinom {n \bmod 2} {k \bmod 2} \pmod 2$