The sum
\[
\frac{1^2-2}{1!} + \frac{2^2-2}{2!} + \frac{3^2-2}{3!} + \cdots + \frac{2021^2 - 2}{2021!}
\]
can be expressed as a rational number \(N\). Find the last 3 digits of \(2021! \cdot N\).

Let's think step by step and output the final answer within \boxed{}.
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