Oxford University MAT 2021 Solved Papers

Let's call the product of the first $\mathit{n}$ terms ${\mathit{b}}_{\mathit{n}}$. Then we have ${\mathit{b}}_{\mathit{n}}={\mathit{a}}_{\mathit{n}}{\mathit{b}}_{\mathit{n}-1}$ (that's how the product works). We also have the definition of ${\mathit{a}}_{\mathit{n}}$ to interpret; it's one more than the previous product, so ${\mathit{a}}_{\mathit{n}}={\mathit{b}}_{\mathit{n}-1}+1$. We can use this to eliminate ${\mathit{b}}_{\mathit{n}}$ and ${\mathit{b}}_{\mathit{n}-1}$ from the previous equation, to get ${\mathit{a}}_{\mathit{n}+1}-1={\mathit{a}}_{\mathit{n}}\left({\mathit{a}}_{\mathit{n}}-1\right).$ Adjust the subscripts and rearrange for ${\mathit{a}}_{\mathit{n}}={\mathit{a}}_{\mathit{n}-1}\left({\mathit{a}}_{\mathit{n}-1}-1\right)+1$