$63^{63^{63}} \mod 100$ I need to find $63^{63^{63}} \bmod 100$. 
This is what I've got so far: 
Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$ 
Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$
Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$
I'm stuck here and don't know what to do next, what could I do now?

You might use Chinese Remainder Theorem.  $100 = 4 \cdot 25$, and
$$63^7 \equiv (-1)^7 \equiv 3 \pmod{4}.$$
Then $$63^7 \equiv 9^7 7^7 \equiv 729\cdot 729 \cdot 9 \cdot 49^3\cdot 7
\equiv 4\cdot 4 \cdot 9 \cdot (-1)^3 \cdot 7 $$
$$\equiv 144(-7) \equiv (-6)(-7) \equiv 42 \pmod{25}.$$
So you just have to add or subtract various multiples of $25$ to $42$ until you get a number that's $3 \pmod{4}.$  And $42+25 = 67$ works.
