$63^{63^{63}} \mod 100$ 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?

$63^7 = (60 +3)^7= ..... + {7\choose 2}60^2\cdot 3^5 + 7\cdot 60\cdot 3^6 + 3^7 \equiv 20\cdot 3^6 +3^7\pmod{100}$.
And as $2\cdot 3^6 \equiv 2\cdot 3^2 \equiv 8 \pmod {10}$ and as $3^7 = 9^3\cdot 3 = (10-1)^3 \cdot 3 = 3(100^3 - 3\cdot 100 + 3\cdot 10 - 1)\equiv 90-3 \equiv 87 \pmod {100}$
So have $63^7\equiv 20\cdot 3^6 + 3^7 \equiv 80 + 87 \equiv 67 \pmod{100}$.
.....
Also nothing wrong with  successive squaring.  $7$ is a small number.
$63^2 = 3969 \equiv 69$.
So $63^3 \equiv 69\cdot 63 = 4347 \equiv 40 + 80 + 27\equiv 47\pmod {100}$.
So $63^6 \equiv 47^2 \equiv 2209\equiv 9\pmod{100}$ and so
$63^7 \equiv 9\cdot 63 \equiv 567 \equiv 67 \pmod {100}$
