Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.

Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$

attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks

For $n\leftarrow 2$, the equality holds.
Let's assume it holds for $n\leftarrow k\in\Bbb{N}$, $k\ge 2$. Then
$$\color{black}{\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 = 2^{2k-1}}$$
We want to show that the equality holds for $k+1$, i.e., that
$$\color{red}{\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k+1}}$$
In fact
$\color{black}{\begin{aligned}
\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 &&=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=2^{k-1}+1}^{2^k}(-1)^i(2i-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^{2^{k-1}+i}(2(2^{k-1}+i)-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^k+2i-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)+(2i-1)^2) &=\\
2\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)) &=\\
2(2^{2k-1}) &+ \underbrace{\sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k})}_0 + \sum_{i=1}^{2^{k-1}}(-1)^i(2^{k+1}(2i-1)) &=\\
2^{2k} &+ 2^{k+1}\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1) &
\end{aligned}}$
It's well known (and easy to show) that $\begin{aligned}\sum_{i=1}^{m}(-1)^i(2i-1) = (-1)^m m\end{aligned}$, so it is proven that
$$\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k} + 2^{k+1}2^{k-1} = 2^{2k+1}$$
