elementary number theory - 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 $77^{17} \equiv x \pmod{100} \Longleftrightarrow \ x= 97$.