Counting odd entries in a row of Pascal's triangle

Idea. The coefficients of $(1+x)^n$ are the binomial coefficients; reducing modulo $2$ turns the expansion into a product of very simple factors.

Frobenius step. Modulo $2$, $(1+x)^2=1+2x+x^2\equiv1+x^2$. Repeating, $(1+x)^{2^i}\equiv(1+x^{2^i})$ for all $i\ge0$.

Factorising. Write $n=\sum_{i\in S}2^i$ with $S$ the set of positions of ones in the binary expansion, so $|S|=s(n)$. Then
$$(1+x)^n=\prod_{i\in S}(1+x)^{2^i}\equiv\prod_{i\in S}\left(1+x^{2^i}\right)=\sum_{T\subseteq S}x^{\sum_{i\in T}2^i}\pmod2.$$

Counting. The exponents $\sum_{i\in T}2^i$ for different subsets $T\subseteq S$ are pairwise different. So the right side is a sum of $2^{|S|}$ distinct monomials with coefficient $1$, and every other coefficient is $0$ modulo $2$. Thus exactly $2^{s(n)}$ coefficients $\binom nk$ are odd.

Check. $n=6=110_2$: row $1,6,15,20,15,6,1$ has $4=2^2$ odd entries.

Remark. In particular the whole row consists of odd numbers iff $n=2^j-1$.

Lesson. Reduce a polynomial identity modulo a prime and use $(1+x)^p\equiv1+x^p$; binary expansions turn $(1+x)^n$ into a product.