Obviously, all terms are odd. We have $a_1^2 < a_1^2+a_1+1 < a_1^2+2a_1+1$ so exists $p \mid a_1^2+a_1+1$ such that $2 \nmid v_p(a_1^2+a_1+1)$. Claim: $a_{n+1} \equiv a_{n} \pmod p$ for $n \geq 2$. Proof: $a_{n+1} \equiv (1+(1+(1+\dots+(1+(1+a_1^2)^2)\dots)^2)^2)^2+1$ with $n$ parentheses. Since $p \mid a_1^2+a+1$, $(1+(1+(1+\dots+(1+(1+a_1^2)^2)\dots)^2)^2)^2+1 \equiv (1+(1+(1+\dots+(1+(-a_1)^2)\dots)^2)^2)^2+1 \equiv (1+(1+(1+\dots+(1+a_1^2)\dots)^2)^2)^2+1 \equiv a_n \pmod p$ with $n-1$ parentheses. Claim: $p \nmid a_2^2+a_2+1$ Notice that $a_2 = a_1^2+1 \equiv -a_1$ so assuming for contradiction we would have $p \mid a_1^2-a_1+1$ so $p \mid 2a_1$ and $p \mid 2a_1^2+2a_1+2$ so $p \mid 2$ but we know $p$ must be odd so we are done. Hence $v_p\left(\prod \limits_{k=1}^N\right) = v_p(a_1^2+a_1+1)$ which is odd, so it is not a perfect square. It is easy to arrive at this hypothesis by experimenting in $p = 3$.