A curious Hankel determinant Define the sequence  $a_{n}$ by $a_{n}=1$ if $n+1=2^k$ for some $k$ and $a_{n}=0$ else.
Computer experiments suggest that the determinant of the Hankel matrix 
$$H_{n+1}:=\begin{pmatrix}
 a_{0} & a_{1} & \dots & a_{n}\\
 a_{1} & a_{2} & \dots & a_{n+1}\\
 \vdots & \vdots & \ddots & \vdots\\
 a_{n} & a_{n+1} & \dots & a_{2n}
      \end{pmatrix}$$
satisfies $$\det{H_{n+1}}=(-1)^\binom{n+1}{2}.$$
Is there a simple way to prove this?
Edit:
Let $H_{n}=(h(i,j)).$ 
For each $n$  there is a unique permutation of ${(0,1,\dots,n-1)}$ such that the determinant of $H_{n}$ equals $h_{0,p(0)}h_{1,p(1)}\dots h_{n-1,p(n-1)}.$
Let me show this in the following example where for clarity I have set $a(n)=x(n)$ if $n+1$ is a power of $2.$
$$H_{9}=\begin{pmatrix}
 x(0) & x(1) & 0 & x(3) & 0 & 0 & 0 & x(7)& 0\\
 x(1) & 0 & x(3) & 0 & 0 & 0 & x(7) & 0& 0\\
    0 & x(3) & 0 & 0 & 0 & x(7) & 0 & 0& 0\\
    x(3) & 0 & 0 & 0 & x(7) & 0 & 0 & 0& 0\\
 0&0&0&x(7)&0&0&0&0&0\\
0&0&x(7)&0&0&0&0&0&0\\0&x(7)&0&0&0&0&0&0&0\\
    x(7) & 0 & 0 & 0 & 0 & 0 & 0 & 0&x(15)\\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & x(15)& 0
      \end{pmatrix}$$
If we  go from right to left we see that $p(8)=7,p(7)=8.$ Then $p(6)=1,p(5)=2,p(4)=3,p(3)=4,p(2)=5,p(1)=6.$ There remains $p(0)=0.$ In general the same procedure works.
Thus my question reduces to a proof of the fact that the sign of this permutation is $(-1)^\binom{n}{2}.$
 A: Edit: following our exchange, I modify the presentation of what is now a partial result.
I will stick to your notations. For example $H_8$ is the $2^3 \times 2^3 $ Hankel matrix:
$$H_{8}:=\begin{pmatrix}
 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1\\
 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\
    0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
    1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
 \vdots & \vdots & \vdots & \vdots &\vdots & \vdots & \vdots & \vdots\\
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0
      \end{pmatrix}$$
We are going to prove a restricted version of your result, i.e., we will prove that, for $k\geq 2$: 
$$\det(H_{2^k})=1$$
complying with formula $(-1)^{\binom{2^k}{2}}=(-1)^{\tfrac{2^k(2^k-1)}{2}}=1.$
Let us set, for notational convenience, $K_n=H_{2^n}$.
$K_n$ possesses a recursive structure with the form : 
$$\tag{1}K_{n+1}= \begin{pmatrix}K_{n} & J_{n}\\J_{n} & 0_{n}\end{pmatrix}$$
where $J_n$ is the anti-diagonal matrix (with ones on the secondary diagonal) and  $0_{n}$ is the all-zero matrix, both of size $2^n \times 2^n$.
Now (1) allows to work by recursion for obtaining $\det(K_n)$ by using the Schur formula for $2 \times 2$ block-defined matrices :
$$M=\begin{pmatrix}A & B\\C & D\end{pmatrix} \ \ \implies \ \ \det(M)=\det(A)det(D-CA^{-1}B)$$
giving here:
$$det(K_{n+1})=\det(K_n)\det(-JK_{n}^{-1}J)=\det(K_n)\det(-I)\det(J)\det(K_{n}^{-1})\det(J)=$$
$$det(-I)det(J_n^2)\det(K_nK_{n}^{-1})=(-1)^{2^n}=\begin{cases}-1 & (n=0)\\+1& (n>0) \end{cases}$$
(where $I$ denotes the identity matrix with $2^n \times 2^n$ elements).
Explanations: $\det(J_n^2)=\det(I)=1$ and $\det(-I_k)=(-1)^k$.

A Matlab program for the recursive generation of matrix $K_{2^p}$ (here with $p=2$):

p=2;
K=[1];J=[1];Z=[0];
for k=1:p
   K=[K,J;J,Z]
   J=[Z,J;J,Z];
   Z=[Z,Z;Z,Z];
end;
K

Edit : I wonder if a simple solution couldn't be achieved by using [Lucas Theorem] (https://en.wikipedia.org/wiki/Lucas%27s_theorem)

