Why it works: Ekadhikena Pūrveṇa

For numbers ending in 5: $n = 1$$0a + 5$. Then $n^2 = (10a+5)$² $= 10$$0a^2+10$$0a+25 = 10$0a($a+1$)+25. The last two digits are always 25; the left part is $a\times (a+1)$ — prefix times one-more-than-prefix. For recurring decimals of $1/(10n+9)$: the multiplier is ($n+1$), and repeatedly multiplying from the right generates the full repeating cycle.