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Here \(h(0) = 1\text{.}\) To get the recurrence relation, think about how you can get \(h(n+1) = (n+1)!\) from \(h(n) = n!\text{.}\) If you write out both of these as products, you see that \((n+1)!\) is just like \(n!\) except you have one more term in the product, an extra \(n+1\text{.}\) So we have,

\begin{equation*} h(0) = 1;~ h(n+1) = (n+1)\cdot h(n)\text{.} \end{equation*}
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