Let
$$ Y := \left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \\ \end{array} \right)^{\otimes k} $$
where $\otimes$ denotes the Kronecker product. We have
$$ \det(Y) = \left(\sqrt{2^k}\right)^{2^k} = 2^{k2^{k-1}} $$
I know how to derive the result by induction, but I have no idea why the first equation is true.
On
Let me give an answer which may explain where the "first equation" comes from.
Note that
$$ \begin{pmatrix} 1 & -1\\ 1& 1\end{pmatrix}= \sqrt{2}\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{pmatrix}= \begin{pmatrix}\sqrt{2} & 0\\0 &\sqrt{2}\end{pmatrix}\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{pmatrix}. $$
The matrix $\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{pmatrix}$ has determinant $1$, and therefore its tensor powers $\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{pmatrix}^{\otimes k}$ also have determinant $1$.
The determinant of $\begin{pmatrix} 1 & -1\\ 1& 1\end{pmatrix}^{\otimes k}$ is therefore equal to the determinant of $\begin{pmatrix}\sqrt{2} & 0\\0 &\sqrt{2}\end{pmatrix}^{\otimes k}$. This latter matrix is a diagonal $2^k \times 2^k$ matrix each of whose entries is $\sqrt{2}\,^k$. Hence the required determinant is $(\sqrt{2}\,^k)^{2^k}$.
I agree that this is not $(\sqrt{2^k)}^{2^k}$; I don't think there's a natural direct way to see that this is true.
Just getting this off the unanswered list.
In general, if $A_i$ is an $n_i\times n_i$ matrix for $i=1,\ldots,k$, then
$$\det \left(\bigotimes_{i=1}^k A_i \right) = \prod_{i=1}^k \det(A_i)^{n_1\cdots n_k/n_i}\tag{1}$$ This can be proved by induction on $k$, where for $k=2$ we can use methods suggested in the comments.
Taking $A_i=A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ for all $i$ in (1), we obtain
$$\det(A^{\otimes k})=\prod_{i=1}^k 2^{2^k/2}=\left(2^{2^{k-1}}\right)^k=2^{k2^{k-1}}\tag{2}$$