Show that $|\det(A_n)|=n^{n/2}$ For k $\ge2$ we recursively define $A_{2^k}$ as $\begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \\ A_{2^{k-1}} & -A_{2^{k-1}} \end{bmatrix}$ and $A_1=[1]$
The problem is to show that $|\det(A_n)|=n^{n/2}$
My attempt: we do an induction on $k$
$|\det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|\det(A_{n})|=n^{n/2}$ and we want to show that $|\det(A_{2n})|=(2n)^n$
using block matrix properties
$|\det(A_{2n})|=|\det(\begin{bmatrix} A_{n} & A_{n} \\ A_{n} & -A_{n} \end{bmatrix})|=|\det(-A)\det(A+AA^{-1}A)|=|2^n\det(A_n)^2|=|2^nn^n|=(2n)^n$
Can somone confirm that there are no flaws in the reasoning please?

Your proof looks good.
Alternatively, for the induction-step, notice that
$$
\begin{vmatrix}
A_n & A_n \\
A_n & -A_n
\end{vmatrix}
=
\begin{vmatrix}
A_n & A_n \\
0 & -2A_n
\end{vmatrix},
$$
since the matrices
$$
\begin{bmatrix}
A_n & A_n \\
A_n & -A_n
\end{bmatrix}
\text{and}
\begin{bmatrix}
A_n & A_n \\
0 & -2A_n
\end{bmatrix}
$$
are row-equivalent.
Thus,
$$|A_{2n}| = |A_n|\cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}.$$
