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?
 A: 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.
A: Not sure that it could be a simpler approach, but an alternative way to show it could be to look at the patterns of the diagonals expressing the $\binom {n}{2^i} $ coefficiens for a given $i $ (that you correctly highlighted by colors in the triangle), and to show that they are identical to the patterns followed by the $i^{th}$ digit (from the right) of the sequence of binary numbers.
For example, it is straightforward to note that the $1^{st} $ digit from the right (i.e. the last) in the sequence of binary numbers follows an alternating pattern $10101010...$ with period $2$, reflecting the behaviour of the progressive increasing numbers $\mod 2$. An identical pattern is followed by the diagonal expressing the $\binom {n}{1} \mod 2 $ coefficients (or equivalently the $\binom {n}{n-1} \mod 2$ coefficients), since they corresponds to the sequence $n \mod 2$.
Generalizing, the $i^{th} $ digit from the right in the sequence of binary numbers follows an alternating pattern with period $2^i$ in which the first $2^{i-1}$ elements are $1$ and the second $2^{i-1}$ elements are $0$. This can be showed by observing that, for any binary number $K $, if the quantity $K \mod 2^i $ is $<2^{i-1} $ then the $i^{th} $ digit from the right must necessary be $0$, whereas if  $K \mod 2^i $ is $\geq 2^{i-1} $ then the $i^{th} $ digit from the right must necessary be $1$. Also, since in the sequence of binary numbers the $i^{th} $ digit from the right compares for the first time at $2^{i-1}$, the sequence starts with $1$. 
A similar pattern exists for the diagonal expressing the $\displaystyle \binom {n}{2^{i-1}} \mod 2 $ coefficients, or equivalently the $\displaystyle \binom {n}{n-2^{i-1}} \mod 2$ coefficients. These diagonals start at $n= 2^{i-1}$ and, in the original Pascal's triangle (non $\mod 2$) have coefficients corresponding to the sequence $\binom {n}{2^{i-1}}$ for given $i$ and increasing $n$. This sequence can be written as 
$$1$$ $$n+1$$ $$\frac { (n+1)(n+2)}{2}$$ $$\frac {(n+1)(n+2)(n+3)}{2 \cdot 3} $$ $$\frac {(n+1)(n+2)(n+3).....(n+j)}{2 \cdot 3 \, ...... \cdot j}$$
and so on. It is not difficult to show that the numerator and the denominator in this sequence have the same divisibility by $2$ (i.e.  their ratio is $\mod 2 = 1$) until in the numerator we reach the $n + 2^{i-1}= 2^i$ term and in the denominator the $2^{i-1}$ term. From this point, the numerator contains the factor $2$ one more time than the denominator, and we have that the ratio is $\mod 2 = 0$. This continues until in the numerator we reach the $n + 2^{i}=2^{i-1}+2^{i}$ term and in the denominator the $2^{i}$ term, where the denominator gets one factor $2$ more than the numerator and we get again that the ratio is $\mod 2=1$, and so on. Such regularity leads to the pattern described above, identical to that of the $i^{th}$ digit from the right in the binary numbers.
A: I'll show how to do this by using the way numbers are written in base $2.$
Fact: Let $f(n)$ be the greatest $k$ such that $2^k$ divides $n!.$ Then, for $n\ge 0,$ $$f(n)=n-(\text{the number of }1\text{s in the binary representation of }n).$$
This fact is a consequence of Legendre's formula; I'll give a standalone proof of this particular instance of Legendre's formula at the end of the answer.  Using the fact, it's easy to conclude that your Pascal's triangle formula is correct, as follows.
I'll write $\text{"the number of }1\text{s in }x\text{"}$ to mean $\text{"the number of }1\text{s in the binary representation of }x\text{".}$
For $2^c \le n,$ the binomial coefficient $\dbinom{n}{2^c}=\dfrac{n!}{(2^c)!\,(n-2^c)!}$ is always an integer, and it will be even or odd according to whether $$f(n)\gt f(2^c)+f(n-2^c)$$ or $$f(n)= f(2^c)+f(n-2^c),$$ respectively.
So \begin{align}\require{cancel}\binom{n}{2^c}\text{ is odd }&\iff f(n)= f(2^c)+f(n-2^c)
\\&\iff \bcancel{n}-(\text{the number of }1\text{s in }n)=
\\&\hphantom{\iff \iff}\cancel{2^c}-(\text{the number of }1\text{s in }2^c)+(\bcancel{n}-\cancel{2^c})-(\text{the number of }1\text{s in }(n-2^c))
\\&\iff (\text{the number of }1\text{s in }n)=1+(\text{the number of }1\text{s in }n-2^c)\scriptsize{\quad\text{(since }2^c\text{ has exactly one }1\text{ in it)}}
\\&\iff(\text{bit #}c\text{ of }n)=1\scriptsize\quad\text{(where bit #}0\text{ is the }1\text{s bit, bit #}1\text{ is the $2\text{s}$ bit, bit #}2\text{ is the $4\text{s}$ bit, etc.).}
\end{align}
That last equivalence is true because:
$\quad\bullet\;$ If bit $\#c$ of $n$ is $1,$ then $n-2^c$ is the same in binary as $n$ except that bit $\#c$ is changed from $1$ to $0.$ So the number of $1\text{s}$ in $n$ is one more than the number of $1\text{s}$ in $n-2^c.$ 
$\quad\bullet\;$ Conversely, if  bit $\#c$ of $n$ is $0,$ find the first $1$ to the left of that $0$ (there must be one since $2^c\le n\text{).}$ That section of $n$ in base $2,$ going from the $1$ we just found up to and including the $0$ at bit $\text{#}c,$ consists of one $1$ followed by one or more $0\text{s}.$  The number $n-2^c$ in base $2$ is the same as $n$ except for that section, which has the $1$ changed to $0$ and all the $0\text{s}$ changed to $1\text{s}.$ It follows that in this case, the number of $1\text{s}$ in $n-2^c$ is greater than or equal to the number of $1\text{s}$ in $n.$
But this equivalence is precisely the statement
$$\binom{n}{2^c}\equiv (\text{bit #}c\text{ of n}) \pmod 2,$$
which is what you wanted.

Let's go back and prove the case of Legendre's formula that we need.  I'll re-state it here:
Fact: Let $f(n)$ be the greatest $k$ such that $2^k$ divides $n!.$  Then, for $n\ge 0,$ $$f(n)=n-(\text{the number of }1\text{s in the base-}2\text{ representation of }n).$$
Proof: We'll use induction on $n.$
The fact is obviously true for $n=0.$
Now assume, as the induction hypothesis, that $f(n)=n-(\text{the number of }1\text{s in the base-}2\text{ representation of }n).$
Since $(n+1)!=n!\,(n+1),$ we have $$f(n+1)=f(n)+\big(\text{the greatest }k\text{ such that }2^k\text{ divides }n+1\big).$$
But the greatest $k$ such that $2^k$ divides $n+1$ is just the number of $0s$ at the end of the $\text{base-}2$ representation of $n+1,$ so we have
\begin{align}f(n+1)&=f(n)+\text{ the number of }0\text{s at the end of }n+1
\\&=n-\text{(the number of }1\text{s in }n\text{) }+\text{ (the number of }0\text{s at the end of }n+1\text{)}. \tag{1}
\end{align}
Now, write $n$ and $n+1$ in base 2.  For convenience, we'll assume that $n$ has at least one $0$ in it by the expedient of prepending a $0$ at the beginning of $n$ if necessary (which doesn't change number of $1\text{s}$ in it.)
You can see that $n$ and $n+1$ written in base 2 are then exactly the same except that if $n$ ends in a $0$ followed by $z\;1\text{s},$ then those $z+1$ bits are replaced in $n+1$ by a $1$ followed by $z\;0\text{s}.$
So:
\begin{align}\text{(the number of }0\text{s at the end of }n+1 \text{)}&=\text{(the number of }1\text{s at the end of }n\text{)}
\end{align}
and
\begin{align}\text{(the number of }1\text{s in }n+1 \text{)}&=\text{(the number of }1\text{s in }n\text{)}-\text{(the number of }1\text{s at the end of }n\text{)}+1.
\end{align}
It follows from $(1)$ above that
\begin{align}
f(n+1)&=n-\text{(the number of }1\text{s in }n\text{) }+\text{(the number of }1\text{s at the end of }n\text{)}
\\&=n-\big(\text{(the number of }1\text{s in }n+1\text{)}-1\big)
\\&=(n+1)-\text{(the number of }1\text{s in }n+1\text{),}
\end{align}
completing the proof.
