Answer

Note that $a_n = \left(\sqrt{1 + \sqrt{a_{n-1}}} - 1\right)^2$ for all $n \ge 2$. Let $b_n = 1 + \sqrt{a_n}$ for all $n \ge 1$, then $b_1 = 2$ and $b_n = \sqrt{b_{n-1}}$ for all $n \ge 2$. So $b_n = 2^{\frac{1}{2^{n-1}}}$ for all $n \ge 1$. Then \begin{align*} S_{2023} &= \sum_{k=1}^{2023} 2^ka_k = \sum_{k=1}^{2023} 2^k(b_k - 1)^2 = \sum_{k=1}^{2023} 2^k(b_k^2 - 2b_k + 1) \\ &= 2 + \sum_{k=2}^{2023} 2^k(b_{k-1} - 2b_k + 1) \\ &= 2 + \sum_{k=2}^{2023} [2^k(b_{k-1} - 1) - 2^{k+1}(b_k - 1)] \\ &= 2 + 4(b_1 - 1) - 2^{2024}(b_{2023} - 1) \\ &= \boxed{6 + 2^{2024} - 2^{2024 + \frac{1}{2^{2022}}}}. \end{align*}