Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p

Theorem
Let $p = 2 n + 1$ be an odd prime.

Then:
 * $F_p \equiv 5^n \pmod p$

Proof
From Fibonacci Number by Power of 2:
 * $2^p F_p = 2 \ds \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k$

From Binomial Coefficient of Prime:
 * $\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$

and from Fermat's Little Theorem: Corollary 1:
 * $2^p \equiv 2 \pmod p$

Hence: