Ok, I finally got this after some hints from math154. Anyways, so first note that the following identity holds: $\sum_{n \ge 0}S(n, k)x^n = x^k\prod_{r = 1}^k\frac{1}{1-rx}$. So \[T(x) = \sum_{n \ge 0}t_nx^n = \sum_{n \ge 0} \sum_{k \ge 0}S(n,k) F_k x^n = \sum_{k \ge 0} \sum_{n \ge 0}S(n,k) F_k x^n = \sum_{k \ge 0} F_kx^k \prod_{r = 1}^k\frac{1}{1-rx} \] Now we'll work in $\mathbb{F}_{p^2}$. Let, $\alpha = \frac{1+\sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2}$, so $F_k = \frac{\alpha^k-\beta^k}{\sqrt{5}}$. For simplicity now, set $P = \prod_{r = 1}^p\frac{1}{1-rx}$. Note that (this is pretty important) $\prod_{r = 1}^{k+p}\frac{1}{1-rx} = P \prod_{r = 1}^k\frac{1}{1-rx}$. We'll continue to simplify $T(x)$. \[\sum_{k \ge 0} F_kx^k \prod_{r = 1}^k\frac{1}{1-rx} = \frac{1}{\sqrt{5}} \sum_{k \ge 0} ((\alpha x)^k-(\beta x)^k) \prod_{r = 1}^k\frac{1}{1-rx} = \frac{1}{\sqrt{5}}\sum_{k = 0}^{p-1}\prod_{r = 1}^k\frac{1}{1-rx}((\alpha x)^k(1+\alpha^p x^p P+\alpha^{2p}x^{2p} P^2+\ldots)-(\beta x)^k(1+\beta^p x^p P+\beta^{2p} x ^{2p} P^2+\ldots) ) \] \[ = \frac{1}{\sqrt{5}}\sum_{k = 0}^{p-1}\prod_{r = 1}^k\frac{1}{1-rx}(\frac{(\alpha x)^k}{1-P\alpha^p x^p}-\frac{(\beta x)^k}{1-P\beta^p x^p})\] Now let $S = \frac{1}{P} = \prod_{r = 1}^p(1-rx) = -x^{p-1}+1$. Write $P$ in terms of $S$ in the above, and simplify. \[ \frac{1}{\sqrt{5}}\sum_{k = 0}^{p-1}\prod_{r = 1}^k\frac{1}{1-rx}(\frac{(\alpha x)^k}{1-P\alpha^p x^p}-\frac{(\beta x)^k}{1-P\beta^p x^p}) = \frac{1}{\sqrt{5}}\sum_{k = 0}^{p-1}S\prod_{r = 1}^k \frac{1}{1-rx}(\frac{(\alpha x)^k}{S-\alpha^p x^p}-\frac{(\beta x)^k}{S-\beta^p x^p}) = \frac{1}{\sqrt{5}}\sum_{k = 0}^{p-1}S\prod_{r = 1}^k \frac{1}{1-rx}(\frac{(\alpha x)^k}{1-x^{p-1}-\alpha^p x^p}-\frac{(\beta x)^k}{1-x^{p-1}-\beta^p x^p}) \] Note that $S\prod_{r = 1}^k \frac{1}{1-rx}$ is a polynomial of finite degree. So we can finish by showing that in $\mathbb{F}_{p^2}$, $T(x) \cdot (x^{p^{2p}-1}-1 )$ is a polynomial (that would mean that all $x^n, x^{n+p^{2p}-1}$ terms would have to cancel each other out, so we'd be done). So it suffices to show that ${1-x^{p-1}-\alpha^p x^p} | x^{p^{2p}-1}-1$. Note that the previous is true iff when we reverse the coefficients it's true, i.e. $x^p - x - \alpha^p | x^{p^{2p}-1}-1$. Note that by the Frobenius endomorphism, if $r$ is a root of $x^p - x - \alpha^p$, then $(r+1)^p - (r+1) - \alpha^p = r^p-r-\alpha^p = 0$, so $r+1$ is a root. So if $x^p - x - \alpha^p$ has a root in $\mathbb{F}_{p^2}$, then it has $p$ roots in that field, so we'd be done. Otherwise assume it doesn't have roots in that field. Then its roots are in some field extension, so by what I previously wrote, all roots are in that extension. Since it has no double roots, it is the product of the distinct minimal polynomials of its roots. But all its roots differ by an integer, so the degree of their minimal polynomials is the same. So the degree divides $p$, so its either $1$, or $p$. The first case is impossible since I said the roots don't lie in $\mathbb{F}_{p^2}$. So the minimal polynomial has degree $p$ so it's irreducible. So it still divides $x^{p^{2p}-1}-1$ (all irreducible polynomials of degree dividing $\frac{2p}{2} = p$ divide that polynomial). For some reason, I couldn't find an easier way to show the divisibility, maybe someone can help me?

I like your approach a lot (it tells us why the period should divide $(p^2)^p - 1$). I had two methods in mind when I wrote the problem: if we let $f(x) = x^p - x - \alpha^p$, then (1) $\alpha^{p^3} = \alpha^p$ gives $f(x)^{p^2} - f(x) = x^{p^3} - x^{p^2} - x^p + x$ (in $\mathbb{F}_{p^2}$), after which we can finish as in the official solution here; and (2) we can note that (again in $\mathbb{F}_{p^2}$) \begin{align*} f(x) &\mid f(x)+f(x)^p+\cdots+f(x)^{p^{2p-1}} \\ &= (x^p-x)+(x^{p^2}-x^p)+\cdots+(x^{p^{2p}} - x^{p^{2p-1}}) - p(\alpha^p+\alpha) \\ &= x^{p^{2p}} - x. \end{align*}