I want to study the parity of $\binom{n}{k}$. I know that there are several ways to do this for a given $n$ and a given $k$, such as Kummer's theorem or Lucas' theorem, which give a method to find the number of factors of $2$ of $\binom{n}{k}$. However, I'm interested in the following problem:
Find all natural numbers $n$ such that $\binom{n+1}{k}$ is even $\forall~k=1,\dots,n$
I know that for $n$ even $\binom{n+1}{1}=n+1$ which is odd, so we can assume that $n$ is odd. It is also easy to prove that in this case $\binom{n+1}{k}$ is even if $k$ is odd. However when $k$ is even $\binom{n+1}{k}$ can be both even or odd, for example $\binom{10}{8}=45$ and $\binom{10}{6}=210$. Checking some cases by hand I was able to see that this is true for $n=3,7,15$ and it falils for $n=5,9,11,13$.
I'm not sure about this, but I guess that the only values that satisfy this property are when $n=2^{i}-1$. Any help on how to prove this fact or any counterexamples at my hypothesis will be appreciated. Thanks!