Pascal's odd rows

Take a look at this figure of Pascal's triangle:

All the even/odd numbers have been shaded differently, and you can see that all the rows $2^n (n\geq 0)$ have all odd numbers, and the others do not. Therefore we have $\boxed{10}$ rows with all odd numbers that are less than $1000$ , namely rows $2^0, 2^1, 2^2, ... 2^9$ .

For those who aren't convinced by the drawing, here is my attempt at a proof:

By Lucas' theorem we know that $\binom{2^k-1}{m}$ is always odd, since $2^k-1$ is all 1s when written in binary. And this corresponds to the rows mentioned above.

Therefore we know that all the rows mentioned above ( $2^0\rightarrow 2^9$ ) only contain odd numbers. Since they contain odd numbers, that means that the row directly underneath them will be all even numbers (except for the two 1's at the end). So, foreach, $n$ , row $2^n + 1$ will contain $2^n -1$ even numbers. This means that the following $2^n-2$ rows will also contain even numbers, which then accounts for all rows. Therefore the rows $2^n$ (for n=0 to 9) are in fact the only rows containing all odd numbers.

QED

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Image credit: http://www.texample.net/