What remains?

$8^{2n} - 62^{2n+1}$ = $64^n - 62^{2n+1} = (63+1)^n - (63-1)^{2n+1}$ ------------ (1)

Applying Binomial expansion for (1) we observe that every term has 63 i.e, is divisible by 9 except

$nCn*1^n - (2n+1)C(2n+1)*(-1)^{2n+1} = 1+1 = 2$

Nice approach, have a look at this by modular arithmetic.

$8\equiv-1(mod9)$ & $62\equiv-1(mod9)$ .

$8^{2n}\equiv(-1)^{2n}\equiv1(mod9)$ & $62^{2n+1}\equiv(-1)^{2n+1}\equiv-1(mod9)$

Subtracting the results we get,

$8^{2n}-62^{2n+1}\equiv1-(-1)\equiv2(mod9)$

Remainder is $\boxed{9}$