Sum of powers divisible by 53 
Show that $1^{26} + 2^{26} + 3^{26} + \cdots + 26^{26} $ is divisible by $53$.

Inspired by another more basic question asked here... I'm interested to see what elegant proofs you come up with (and mine is also given below in an answer).

Let $ S $ be the given sum, and fix $ \alpha \in \mathbb F_{53}^{\times} $ such that $ \alpha $ is not a root of $ X^{26} - 1 $. Then, (the following equalities are in $ \mathbb F_{53} $)
$$ 2S = \sum_{k=1}^{52} k^{26} = \sum_{k=1}^{52} (\alpha k)^{26} = \alpha^{26} \cdot  2S $$
(Note that $ x \to \alpha x $ is a permutation of $ \mathbb F_{53} $.) It follows that either $ 1 - \alpha^{26} = 0 $ or $ S = 0 $. Since $ \alpha $ was chosen such that the former is not true, the latter must be true; i.e $ S \equiv 0 \pmod{53} $.
