Remainder

$2222\equiv 3 (mod 7)\\ 2222^2\equiv 9\equiv 2 (mod 7)\\ 2222^3\equiv 6 (mod 7)\\ 2222^4\equiv 18\equiv 4 (mod 7)\\ 2222^5\equiv 12\equiv 5 (mod 7)\\ 2222^6\equiv 15\equiv 1(mod 7)$

The pattern repeats after this. Since $5|5555$ , $2222^{5555}\equiv 2222^5\equiv 5(mod 7)$

${2222}^{5555} \pmod7 \equiv 3^{5555} \pmod7$

As $3$ and $7$ are co-prime, we can say that ( By Euler's Theorem )

$3^{\phi(7)} \equiv 1 \pmod7$

$\therefore 3^{5555} \equiv (3^{6 \cdot 925}) \cdot 3^5 \equiv 1 \cdot 243 \equiv \boxed{5} \pmod7$