7. Given a non-zero sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=3, a_{n}\left(a_{n+2}-1\right)=a_{n+1}\left(a_{n+1}-1\right), n \in \mathbf{N}^{*}$. Then the value of $\mathrm{C}_{2023}^{0} a_{1}+\mathrm{C}_{2023}^{1} a_{2}+\mathrm{C}_{2023}^{2} a_{3}+\cdots+\mathrm{C}_{2023}^{2023} a_{2024}$ is
. (Express the result using integer exponents)
$$
\begin{array}{l}
a_{n}\left(a_{n+2}-1\right)=a_{n+1}\left(a_{n+1}-1\right) \Rightarrow \frac{a_{n+2}-1}{a_{n+1}}=\frac{a_{n+1}-1}{a_{n}}=\cdots=\frac{a_{2}-1}{a_{1}}=2, \\
\text { then } a_{n+1}=2 a_{n}+1 \Rightarrow a_{n+1}+1=2\left(a_{n}+1\right) \Rightarrow a_{n}+1=2^{n} \Rightarrow a_{n}=2^{n}-1 . \\
\text { Therefore } \mathrm{C}_{2023}^{0} a_{1}+\mathrm{C}_{2023}^{1} a_{2}+\mathrm{C}_{2023}^{2} a_{3}+\cdots+\mathrm{C}_{2022}^{2023} a_{2024} \\
=2 \mathrm{C}_{2023}^{0}+2^{2} \mathrm{C}_{2023}^{1}+2^{3} \mathrm{C}_{2023}^{2}+\cdots+2^{2024} \mathrm{C}_{2023}^{2023}-\left(\mathrm{C}_{2023}^{0}+\mathrm{C}_{2023}^{1}+\cdots+\mathrm{C}_{2023}^{2023}\right) \\
=2(2+1)^{2023}-2^{2023}=2 \cdot 3^{2023}-2^{2023} .
\end{array}
$$