what is and how can i calculate this congruence? $5^{625} \equiv x \pmod{41}$

Question

$$5^{625} \equiv x \pmod{41}$$

I know that it is like $41 \mid 5^{625}-x$, but I don't know how to continue with that or how can I try it

Answer 1

$5^{625}\equiv5^{25}\bmod41$ by Fermat's little theorem.

$5^3\equiv2\bmod 41$ and $2^7\equiv5\bmod 41$.

Therefore, $5^{25}\equiv2^85\equiv50\equiv9\bmod 41$.