Last 4

$25^{3} \equiv 5625 \pmod{10000}$
$25^{4} \equiv 0625 \pmod{10000}$
$25^{5} \equiv 5625 \pmod{10000}$
$25^{6} \equiv 0625 \pmod{10000}$
Thus,

$25^{2n} \equiv 0625 \pmod{10000} , n \ge 2$
$25^{2n-1} \equiv 5625 \pmod{10000} , n \ge 2$
$\therefore 25^{2001} \equiv 5625 \pmod{10000}$

Yeah, I got the answer, but how can we prove that this is not a coincidence without actually trying?