There seems to be a typo in the inequality. In its current form, is is trivial
(see Servaes' answer). We are going to show a similar and stronger inequality instead.
For any $n \ge 2$ and $a_1, \ldots, a_n \ge 1$, we have
$$\left(\prod_{k=1}^n a_k\right)^2 \color{red}{-} \left(\prod_{k=1}^n a_k\right) + 1
\ge \prod_{k=1}^n (a_k^2 - a_k + 1)
$$
Let $f(x) = x^2 - x + 1$, above inequality can be rewritten as
$$f\left(\prod_{k=1}^n a_k\right) \ge \prod_{k=1}^n f(a_k)
\quad\text{ for } a_1,\ldots,a_n \ge 1
$$
For any $n \ge 2$, let $\mathcal{S}_n$ be the statement that above inequality is true at that paricular $n$.
Notice for any $a, b \ge 1$,
$$\begin{align} f(ab)-f(a)f(b) &= (ab)^2 - ab + 1 - (a^2-a+1)(b^2+b+1)\\
&= (a-1)(b-1)(a+b) \ge 0\end{align}$$
The base statement $\mathcal{S}_2$ is true.
Assume $\mathcal{S}_n$ is true for a particular $n$. For any $a_1,\ldots,a_{n+1} \ge 1$, we have
$$f\left(\prod_{k=1}^{n+1} a_k\right)
= f\left(a_1\prod_{k=2}^{n+1} a_k\right)
\stackrel{\mathcal{S}_2}{\ge} f(a_1)f\left(\prod_{k=2}^{n+1} a_k\right)
\stackrel{\mathcal{S}_n, f(a_1) \ge 0}{\ge} f(a_1)\prod_{k=2}^{n+1}f(a_k)
= \prod_{k=1}^{n+1}f(a_k)
$$
This means $\mathcal{S}_n \implies \mathcal{S}_{n+1}$. By induction, $\mathcal{S}_n$ is true for all $n \ge 2$.
Apply this to the inequality at hand. For any $a_1,\ldots, a_s \ge 2$, we have
$$\begin{align}
\left(\prod_{k=1}^s a_k\right)^2 + \left(\prod_{k=1}^s a_k\right) + 1
> &\left(\prod_{k=1}^s a_k\right)^2 - \left(\prod_{k=1}^s a_k\right) + 1 \\
\ge & \prod_{k=1}^s(a_k^2-a_k+1)\\
> & \prod_{k=1}^s(a_k^2-a_k+1) - 1
\end{align}
$$