Do you even Lift the Exponent?

Notice how $n$ does not have to be an integer. Therefore for those of you who used LTE... you did all that for nothing.

The greatest possible $n$ is when $3^n = 8014^{100!} - 8011^{100!}$ Therefore we need to find this in modulo 1000.

$8014^{100!} - 8011^{100!} \equiv 14^{100!} - 11^{100!} \text{ (mod 1000)}$

Since $gcd(1000, 11) = 1$ , and $\phi(1000) = 400$ , this implies that $11^{100!} \equiv 1 \text{ (mod 1000)}$ .

We just need to fine $14^{100!}$ in mod 1000. To do this, we use the Chinese Remainder Theorem, splitting mod 1000 to mod 8 and mod 125.

Obviously $14^{100!} \equiv 0 \text{ (mod 8)}$ .

Since $\phi(125) = 100 \implies 14^{100!} \equiv 1 \text{ (mod 125)}$

By CRT, we get that $14^{100!} \equiv 376 \text{ (mod 1000)}$

This implies that $14^{100!} - 11^{100!} \equiv 376 - 1 \equiv \boxed{375}$