Verification on proof (If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.) Here is the proposition I'm assessing:
If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
While I've seen many valid answers, 
I really like this answer: https://math.stackexchange.com/a/1002167/367034
However, can someone clarify for me the validity of the following claim within the answer cited above:
Now, if $k$ is either odd or even, $\displaystyle\frac{k+1}{2^n-k}$ is odd.
I understand there are stipulations on the values of $2^n$ and $k$ (i.e. $k\ne 0,2^n$) but I've been able to derive both even simplifications of this expression as well as some values for $k$ that produce non-integer values. Can someone let me know if I'm missing something here?
 A: It seems you are correct: For small values of $k$, $\frac{k+1}{2^n-k}$ is not an integer. Let's try to fix this and show that $\binom{2^n}{k}$ is always even directly (the rest of the argument in the linked answer is fine). We have:
$$\binom{2^n}{k}=\frac{2^n!}{k!(2^n-k)!}=\frac{2^n(2^n-1)\cdots(2^n-(k-1))}{k!}=\frac{2^n}{k}\prod_{j=1}^{k-1}\frac{2^n-j}{j}$$
Let's look at how many $2$s appear in the prime decomposition of the denominator and the numerator:
The power of $2$ in the factorization of $2^n-j$ is the same as the power of $2$ in the factorization of $j$. Thus all the $2$s that appear in the numerator, from terms $2^n-j$, cancel out with all the $2$s that appear in the denominator, from terms $j$, for $1\leq j\leq k-1$. Since we are assuming $k<2^n$ then the power of $2$ in $k$ will be strictly smaller than $n$.
Therefore, there will be some $2$s left after simplifying the fraction, and so $\binom{2^n}{k}$ is even.
A: About the first property you mention, there is a short intuitive explanation (I don't say a true "proof").
We are looking at lines with all their coefficients equal to 1 in Pascal triangle modulo 2. 
But Pascal's triangle modulo 2 is known to modelize in certain way the fractal known as Sierpinki triangle (http://mathworld.wolfram.com/SierpinskiSieve.html).
In this structure, lines where all coefficients are ones are found by the (self-repeating) homothety with ratio 2 that corresponds to the doubling $2^k \rightarrow 2^{k+1}$ if the lines had been numbered beginning at 1 instead of zeros.
