What are the last two digits of $139^{139^{100}}$?
I know that I have to utilize (mod 100) for the problem.
Getting use of Euler's theorem as $(139, 100) = 1$ yields
$$139^{{139}^{100}} \equiv 139^{{139}^{100} \pmod{\Phi(100)}} \pmod{100}$$
$$ \equiv 139^{{139}^{100} \pmod{40}} \pmod{100}.$$
How to proceed from this point?
You want to evaluate $139^{100}\bmod{40}$. Note that $139\equiv19\bmod{40}$.
Finally, since $19=20-1$, $19^2=20^2-2\cdot20+1$,
$19^2\equiv1\bmod{40}$.
Thus, $$139^{19^{100}}\equiv139^1\equiv39\bmod{100}$$