Determination of the last two digits of $777^{777}$ May I know if my proof is correct? Thank you. 
This is equivalent to finding $x$ such that $777^{777} \equiv x  \pmod{100}.$
By Euler's theorem, $777^{\ \psi(100)} =777^{\ 40}\equiv 1 \pmod{100}$.
It follows that $777^{760} \equiv 1  \pmod{100}$ and  $777^{\ 17} \equiv x \pmod{100}.$
By Binomial expansion, $777^{\ 17} = 77^{\ 17}+700m$, for some positive integer $m$.
Hence $77^{17} \equiv x \pmod{100} \Longleftrightarrow \ x= 97$.

Yep! Your proof is correct. I particularly like your use of the binomial expansion to make the terms easier to deal with.
