Combinatorial Proof
Let $n\in\mathbb{N}\cup \{0\}$. First, it is obvious that, if $\dbinom{n}{k}$ is odd for all $k=0,1,2,\ldots,n$, then $n$ is odd. Suppose now that $n=2m-1$ for some integer $m$. The involution $\tau$ on $[n]:=\{1,2,\ldots,n\}$ is defined by $\tau(j)=n+1-j$ for all $j=1,2,\ldots,n$. Then, $\tau_k$ is also an involution on $\dbinom{[n]}{k}=\big\{X\subseteq[n]\,\big|\,|X|=k\big\}$, where $\tau_k(X):=\big\{\tau(x)\,|\,x\in X\big\}$ for all $X\in\dbinom{[n]}{k}$. Note that $\binom{n}{k}\equiv \text{Fix}\left(\tau_k\right)\pmod{2}\,,$ where $\text{Fix}(f)$ is the number of fixed points of $f:[n]\to[n]$. It is evident that $\text{Fix}\left(\tau_k\right)=\binom{m-1}{\left\lfloor k/2\right\rfloor}$ for all $k=0,1,2,\ldots,n$. The claim follows by induction.
There may be a similar combinatorial proof if $2$ is replaced by an odd prime natural number $p$. I would love to see it.