Are the $a_i$ supposed to be real numbers? If so:

CantonMathGuy wrote: Are the $a_i$ supposed to be real numbers? If so:

Yes! Sorry that I forgot it.

Nice and easy. We have $n \geq 2 \in \mathbb{N},a_{i} \in \mathbb{R}$ and $a_i\neq -1,a_{i+2}=\frac{a_i(a_i+1)}{a_{i+1}+1}$ $\forall$ $1 \leq i \leq n$ . If $a_{1}=0$ then $a_{i}=0$ $\forall$ $1 \leq i \leq n$. Thus we assume $a_{1} \neq 0$ throughout. WLOG assume that, $a_{1} \geq a_{2}$. Then, $$a_{3}=\frac{a_{1}(a_{1}+1)}{a_{2}+1} \geq a_{1} \geq a_{2}$$$$a_{4}=\frac{a_{2}(a_{2}+1)}{a_{3}+1} \leq a_{2}$$$$a_{5}=\frac{a_{3}(a_{3}+1)}{a_{4}+1} \geq a_{3}$$So we have the following inequality: (Let $n=2k$ for some $k \in \mathbb{N}$) $$ a_{2k-1} \geq \cdots a_{5} \geq a_{3} \geq a_{1} \geq a_{2} \geq a_{4} \geq a_{6} \cdots \geq a_{2k}$$But we also have, $a_{1}=\frac{a_{2k-1}(a_{2k-1}+1)}{a_{2k}+1} \geq a_{2k-1}$ and $a_{2}=\frac{a_{2k}(a_{2k}+1)}{a_{1}+1} \leq a_{2k}$ i.e, $$\boxed{ a_{1} \geq a_{2k-1} \geq \cdots a_{5} \geq a_{3} \geq a_{1} \geq a_{2} \geq a_{4} \geq a_{6} \cdots \geq a_{2k}\geq a_{2}}$$Now, $a_{3}=a_{1}=\frac{a_{1}(a_{1}+1)}{a_{2}+1}\implies a_{1}=a_{2}$. Hence $\boxed{a_{1}=a_{2}= \cdots =a_{n}}$. Similar analogy for odd $n$ follows. $\blacksquare$ $(\mathbb{QED})$