Proving that when $n$ is not a power of 2, $\exists k, 0 < k < n$ such that $\binom{n}{k}$ is odd So I had a read over this question here which showed that any $n$ that is a power of two has it that $\binom{n}{k}$ is an even number for $0 < k < n$. 
Now I am wondering if for every other $n$, there exists some $k,0 < k < n$ such that $\binom{n}{k}$ is odd.
I was able to prove that every odd numbered $n$ satisfies this statement by saying that when $k=1, \binom{n}{k}$ is odd. However, I am not sure how to prove the other case where $n = 2^pr$ for $p$ being a non-negative integer and $2^p$ and $r$ are relatively prime. Could I have some help?
 A: If $k=2^p$ then $\binom{n}{k}$ will be odd, for $$
\binom{r2^p}{2^p}=\prod_{j=0}^{2^p-1}\frac{r2^p-j}{2^p-j}$$ aand we see that any factor of $2$ in the numerator will be cancelled by a factor of $2$ in the denominator.
A: Lemma: For all nonnegative integers $n,k$, $\binom{2n}{2k}$ and $\binom{n}k$ have the same parity.
Proof: We will take all of the subsets of $\{1,\dots,2n\}$ of cardinality $2k$ and divide most of them into pairs as follows:

For each subset $S$, find the smallest number $i$ so that $i\le n$ and $S$ contains exactly one of the two elements $2i-1$ and $2i$. The mate of $S$ is obtained by exchanging $2i-1$ for $2i$. If no such $i$ exists, then $S$ is unpaired.

Note that the unpaired subsets will contain both or neither of $\{2i-1,2i\}$ for each $i\in \{1,\dots,n\}$. These unpaired sets are obviously in bijection with the subsets of $\{1,\dots,{n}\}$ of size $k$. This proves that
$$
\binom{2n}{2k}\equiv\binom{n}{k}\pmod 2
$$
since $\binom{n}k-\binom{n/2}{k/2}$ is equal to the number of sets in pairs, which is even.
Now, in your case, writing $k=r2^p$, we can repeatedly apply the Lemma to show
$$
\binom{r2^p}{2^p}\equiv \binom{r2^{p-1}}{2^{p-1}}\equiv \dots \equiv \binom{r}1=r\equiv 1\pmod 2
$$
since $r$ is odd. This shows that $k=2^p$ works.
