Find the last two digits of $7^{7^{7^{10217}}}$.  We need to find $7^{7^{7^{10217}}}$ (mod $100$) and  $\phi(100) = 40$ So $7^{40} \equiv 1$(mod$100$) I don't know how to proceed after this
Find the last $2$ digits of $$\large7^{7^{7^{10217}}}$$   So far I have: $17\cdot 601=10217$ and $7=7 \pmod{10}$. Any help greatly appreciated.
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