Pascal's Triangle and Binary Representations In the article that I am currently reading, it is stated as a well-known fact that positions $2^i$ or equivalently $(n-2^i)$ in the $n^{th}$ row in Pascal's Triangle modulo $2$ spell out the binary representation of $n$:
$$
\newcommand{\red}{\color{red}}
\newcommand{\blue}{\color{blue}}
\begin{array}{rrc}
0:&0&1\\
1:&1&\red1\ 1\\
2:&10&\blue1\ \red0\ 1\\
3:&11&1\ \blue1\ \red1\ 1\\
4:&100&\red1\ 0\ \blue0\ \red0\ 1\\
5:&101&1\ \red1\ 0\ \blue0\ \red1\ 1\\
6:&110&1\ 0\ \red1\ 0\ \blue1\ \red0\ 1\\
7:&111&1\ 1\ 1\ \red1\ 1\ \blue1\ \red1\ 1\\
8:&1000&\blue1\ 0\ 0\ 0\ \red0\ 0\ \blue0\ \red0\ 1\\
9:&1001&1\ \blue1\ 0\ 0\ 0\ \red0\ 0\ \blue0\ \red1\ 1\\
10:&1010&1\ 0\ \blue1\ 0\ 0\ 0\ \red0\ 0\ \blue1\ \red0\ 1
\end{array}
$$
To be more precise and/or technical, if $n$'s binary expansion is $b_t b_{t-1}\cdots b_1 b_0$ or equivalently
$
n=\sum_{i=0}^t b_i\cdot 2^i
$
then we have
$$
b_i=\binom n{2^i}\pmod 2
$$

Now I was thinking about the cleanest and simplest way to prove this result. I find the self-similarity of Sierpinski's Triangle to provide a nice visual argument, yet it fails to be simple to communicate succintly in a paper, I think.
My suggested proof
Thus I thought it would be simpler to consider which powers of $2$ divide respectively the numerator and denominator of
$$
\binom n{2^i}=\frac{\prod_{s=1}^{2^i}(n-2^i+s)}{\prod_{t=1}^{2^i} t}
$$
Now note that the factors of the numerator $n-2^i+s$ cover a full set of residues modulo $2^i$. Those with non-zero remainder $n-2^i+s\equiv t$ modulo $2^i$ will have the same divisibility by $2$ as $t$ has, namely some power $2^j<2^i$ will be the maximal power of $2$ dividing both $t$ and that factor.
Exactly one factor will be divisible by $2^i$, namely the single factor $n-2^i+s$ whose binary representation ends in $i$ zeros. Now if the $i^{th}$ bit is zero then this factor will at least be divisible by $2^{i+1}$ because then it ends in at least $i+1$ zeros. If on the other hand the $i^{th}$ bit is $1$, then this factor is divisible by no higher power of $2$ than $2^i$.
Thus we see that if the $i^{th}$ bit is $1$ there is a 1:1 correspondance between the factors of the numerator and denominator with respect to their divisibility by $2$ thus resulting in an odd number. But if the $i^{th}$ bit is zero then the numerator has at least one more factor $2$ than the numerator - counted by multiplicity.

Question: Do you have suggestions to simplify this argument or can you point me to a completely different approach making everything simpler? Perhaps there even is a clean and simple combinatorial proof?

Another approach uses generating functions (for a similar example, see the proof of Lucas's Theorem). Let $p(x) = \sum_{k=0}^n\binom{n}{k}x^k$.
It is easy to check that for primes $p$ and nonnegative integers $k$, we have $(1+x)^{p^k}\equiv 1 + x^{p^k}\pmod p$.
Then
$$p(x) = (1+x)^n = \prod_{i=0}^t \left((1+x)^{2^i}\right)^{b_i} \equiv \prod_{i=0}^t \left(1+x^{2^i}\right)^{b_i}\pmod 2.$$
Thus $\binom{n}{2^j}$ is congruent to the coefficient of $x^{2^j}$ in $\prod_{i=0}^t \left(1+x^{2^i}\right)^{b_i}$ mod $2$.
Since all the $b_i$ are 0 or 1, the coefficient of $x^{k}$ in $\prod_{i=0}^t \left(1+x^{2^i}\right)^{b_i}$ is the number of ways to write $k=2^{i_1}+2^{i_2}+\cdots+2^{i_m}$ for some $i_1<i_2<\cdots<i_m$ where $b_{i_1}=b_{i_2}=\cdots=b_{i_m} = 1$.
Since binary representation is unique, all the coefficients of $\prod_{i=0}^t \left(1+x^{2^i}\right)^{b_i}$ are 0 or 1.
In particular, the coefficient of $x^{2^j}$ is 1 if $b_j=1$ and 0 if $b_j=0$, so we have $b_j\equiv \binom{n}{2^j}\pmod 2$.
I believe by the same argument you can show for all primes $p$, writing $n=\overline{b_tb_{t-1}\dots b_0}_p$ in base $p$, we have
$$b_j\equiv\binom{n}{p^j}\pmod p. $$
EDIT: For this problem and the problem for general $p$ you can actually can just apply Lucas's Theorem directly:
$$\binom{n}{p^j} \equiv \prod_{i=0}^t\binom{b_i}{[i=j]}\equiv b_j\pmod p$$
where we denote $[i=j]$ to be 1 if $i=j$ and 0 otherwise.
