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?

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.
