$7^{7^{7^{7}}}!$

$\phi(100) = 40$

$\phi(40)=16$

We divide the original problem into smaller chunks:

I) $7^{7^{7^{7}}} \pmod{100} \equiv 7^{7^{7^{7} \pmod{16}} \pmod{40}} \pmod{100}$

I) a) $7^7 = 49^3 * 7 \equiv 1^3 * 7 \equiv 7 \pmod{16}$

I) b) $7^7 \equiv 49^3 * 7 \equiv 23 \pmod{40}$

II) $7^{7^{7}} \pmod{100} \equiv 7^{7^{7} \pmod{40}} \pmod {100}$ ; From earlier we know that $7^7 \equiv 23 \pmod{40}$ , so $7^{7^{7}} \equiv 7^{23}$

III) $7^7 \equiv 43 \pmod{100}$

And finally we get:

$7^{23} - 7^{23} + 43 - 7 \equiv \boxed{36} \pmod{100}$