First observe that $p=2$ does not satisfy the condtion, so $p$ must be an odd prime. Lemma. If $p>2$ is a prime and $\mathbb{F}_{p}$ is the field containing $p$ elements, then for any integer $1 \leq n<p$ one has the following equality in the field $\mathbb{F}_{p}$ $$\prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(1+\alpha^{n}\right)= \begin{cases}0, & \text { if } \frac{p-1}{\operatorname{gcd}(p-1, n)} \text { is even } \\ 2^{n}, & \text { otherwise }\end{cases}$$ Proof. We may safely assume that $n \mid p-1$ since it can be easily proved that the set of $n$-th powers of the elements of $\mathbb{F}_{p}^{*}$ coincides with the set of $\operatorname{gcd}(p-1, n)$-th powers of the same elements. Assume that $t_{1}, t_{2}, \ldots, t_{n}$ are the roots of the polynomial $t^{n}+1 \in \mathbb{F}_{p}[x]$ in some extension of the field $\mathbb{F}_{p}$. It follows that $$\prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(1+\alpha^{n}\right)=\prod_{\alpha \in \mathbb{F}_{p}^{*}} \prod_{i=1}^{n}\left(\alpha-t_{i}\right)=\prod_{i=1}^{n} \prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(t_{i}-\alpha\right)=\prod_{i=1}^{n} \Phi\left(t_{i}\right)$$ where we define $\Phi(t)=\prod_{\alpha \in \mathbb{F}_{p}^{*}}(t-\alpha)=t^{p-1}-1$. Therefore $$\prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(1+\alpha^{n}\right)=\prod_{i=1}^{n}\left(t_{i}^{p-1}-1\right)=\prod_{i=1}^{n}\left(\left(t_{i}^{n}\right)^{\frac{p-1}{n}}-1\right)=\prod_{i=1}^{n}\left((-1)^{\frac{p-1}{n}}-1\right)= \begin{cases}0, & \text { if } \frac{p-1}{n} \text { is even } \\ 2^{n}, & \text { otherwise }\end{cases}$$ Let us now get back to our problem. Suppose the numbers $i^{k}+i, 0 \leq i \leq p-1$ form a complete residue system modulo $p$. It follows that $$\prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(\alpha^{k}+\alpha\right)=\prod_{\alpha \in \mathbb{F}_{p}^{*}} \alpha$$ so that $\prod_{\alpha \in \mathbb{F}_{p}^{*}}\left(\alpha^{k-1}+1\right)=1$ in $\mathbb{F}_{p}$. According to the Lemma, this means that $2^{k-1}=1$ in $\mathbb{F}_{p}$, or equivalently, that $2^{k-1} \equiv 1(\bmod p)$. Therefore $a_{2}=2^{k}+2 \equiv 4(\bmod p)$ so that the remainder of $a_{2}$ upon division by $p$ is either 4 when $p>3$ or is 1 , when $p=3$.