Determinants of products of binary matrices and binomial coefficients Consider two binary semi-infinite matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0 &\cdots\\
1 &0 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &0 &0 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
1 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &1 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &1 &0 &0 &\cdots\\
0 &0 &0 &0 &1 &1 &0 &\cdots\\
0 &0 &0 &0 &0 &1 &1 &\cdots\\
0 &0 &0 &0 &0 &0 &1 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Let $A_n=T^n C$, then the non-zero entries of $A_n$ are the ${n+1}\choose{m}$, $0\le m \le n+1$ binomial coefficients. For example, 
$$
A_3=
\begin{bmatrix}
4 &4 &0 &0 &0 &0 &\cdots\\
1 &6 &1 &0 &0 &0 &\cdots\\
0 &4 &4 &0 &0 &0 &\cdots\\
0 &1 &6 &1 &0 &0 &\cdots\\
0 &0 &4 &4 &0 &0 &\cdots\\
0 &0 &1 &6 &1 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Is there a simple formula for the determinants of the $k\times k$ upper left blocks of the matrices $A_n$? That is: what is $\det(B_n^k)$, where $$B_n^k(ij)=A_n(ij),$$ $1\le i,j \le k\le n+1$?
NOTES:


*

*The computer factorisation of $\det(B_n^k)$ shows that
$\det(B_n^{(n+1)}=2^{n(n+1)/2}$ and the determinants factorisations
have only factors less than $2(n+1)$, which suggests that the
determinants are products of binomial coefficients of the form
${l}\choose{k}$, $0\le k,l\le 2(n+1)$.

*This question is motivated by the continued fraction approximation of
the square root function, the matrices $A_n$ being the Hurwitz
matrices of the continued fractions.

It follows from the answer to Coefficients of binomial continued fractions, since the matrices in the question are Hurwitz matrices of the continued fractions that:
$$ \det B_n^1=n,$$
$$ \det B_n^2=\frac{n(n^2-1)}{3},$$
$$ \det B_n^3=\frac{n^2(n^2-1)(n^2-2^2)}{3^2\cdot5},$$
$$ \det B_n^4=\frac{n^2(n^2-1)^2(n^2-2^2)(n^2-3^2)}{3^3\cdot5^2\cdot7},$$
$$ \det B_n^5=\frac{n^3(n^2-1)^2(n^2-2^2)^2(n^2-3^2)(n^2-4^2)}{3^4\cdot5^3\cdot7^2\cdot9},$$
$$ \dots$$
$$ \det B_n^l=n^{\lceil l/2\rceil}\prod_{k=1}^{l-1}\frac{(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}.$$
The fact that $$\det B_n^n=2^{n(n-1)/2}$$ follows from Rational fraction expression for triangular powers of 2
