Find the last two digits of $1717^{1717}$ 
$1717^{1717} \mod 100$

Since $\phi(100) = 40$ , we can transform this into:

$17^{37}101^{37} \mod 100 = 17^{37} \mod 100$

How do I proceed further?

Note that $$17^{37}\equiv(17^3)^{12}\cdot17\equiv(13^{3})^4\cdot17\equiv(-3)^4\cdot17\equiv1377\equiv77\pmod{100}.$$
or
$$17^{37}\equiv(17^4)^9\cdot17\equiv21^5\cdot21^4\cdot17\equiv1\cdot81\cdot17\equiv77\pmod{100}.$$
