show this $ t_{n+p^{2p}-1} \equiv t_n \pmod{p} $ with hard problem 
We define the Fibonacci sequence $\{F_n\}_{n\ge0}$ by $F_0=0$, $F_1=1$, and for $n\ge2$, $F_n=F_{n-1}+F_{n-2}$; we define the Stirling number of the second kind $S(n,k)$ as the number of ways to partition a set of $n\ge1$ distinguishable elements into $k\ge1$ indistinguishable nonempty subsets.
        For every positive integer $n$, let $$t_n = \sum_{k=1}^{n} S(n,k) F_k$$ Let $p\ge7$ be a prime. Prove that $$ t_{n+p^{2p}-1} \equiv t_n \pmod{p} $$ for all $n\ge1$.

some My try: 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 $$T(x) = \sum_{n \ge 0}t_nx^n = \sum_{n \ge 0} \sum_{k \ge 0}S(n,k) F_k x^n = \sum_{k \ge 0} \sum_{n \ge 0}S(n,k) F_k x^n = \sum_{k \ge 0} F_kx^k \prod_{r = 1}^k\frac{1}{1-rx} $$
Then I can't,Thanks
 A: Your sequence is A263576 in OEIS. Since the Fibonacci numbers have the following closed form: 
$$F_n=\frac{\phi^n-(1-\phi)^{n}}{\sqrt{5}}$$
We must have:
$$t_n = \sum_{k=1}^{n} S(n,k) F_k = \sum_{k=1}^{n} S(n,k)\frac{\phi^k-(1-\phi)^{k}}{\sqrt{5}}\\ = \frac{1}{\sqrt{5}}\sum_{k=1}^{n} S(n,k)\phi^k - \frac{1}{\sqrt{5}}\sum_{k=1}^{n} S(n,k)(1-\phi)^k=\frac{B_n(\phi)-B_n(1-\phi)}{\sqrt{5}}$$
Where
$$B_n(x)=\sum_{k=1}^{n} S(n,k) x^k$$
Is the $n$-th Bell polynomial. Now, by Proposition 3.1. of this paper: http://www.kurims.kyoto-u.ac.jp/EMIS/journals/BBMS/Bulletin/bul964/Robert-Gertsch.pdf we obtain: 
$$B_{n+p^{2p}-1}(x) \equiv B_n(x)+(x^p+x^{p^2}+...+x^{p^{2p}})B_n(x) \equiv B_n(x)+(x+...+x)B_n(x) \equiv B_n(x) \pmod{p}$$
So by the formula above we naturally get the result. However for this to work it seems that $\sqrt{5}$ needs to exist mod $p$ which happens only if $p \equiv 1$, $4 \pmod{5}$. There is probably some way to make it work for $p \equiv 2$, $3 \pmod{5}$ I just don't see how. 
An alternative approach would be to make use of Corollary 2.2 from http://matwbn.icm.edu.pl/ksiazki/aa/aa94/aa9413.pdf and known behavior of Fibonacci numbers mod $p$ but I haven't been able to make it work either. 
