The Two Digit..

I'm pretty sure there's a simpler way, but here's how I did it (using the multiplication property of congruences). $9^2 = 81$ (mod $100$ ), $9^3 = 29$ (mod $100$ ), $9^4 = 61$ (mod $100$ ), $9^8 = 21$ (mod $100$ ), $9^9 = 89$ (mod $100$ ), and $9^{10} = 1$ (mod $100$ ). Since the exponent $9^9$ can be written as $100n+89$ for some int. $n$ , $9^{9^9} = 9^{100n+89}=(9^{10})^{10n}9^{89}=(1)^{10n}(9^{10})^89^9=(1)(1)^8(89)=89$ (mod 100).