elementary number theory - The last two digits of $9^{9^9}$

I tried to calculate the last two digits of $9^{9^9}$ using Euler's Totient theorem, what I got is that it is same as the last two digits of $9^9$.

How do I proceed further?

Answer

At this point, it would seem to me the easiest thing to do is just do $9^9 \mod 100$ by hand. The computation should only take a few minutes. In particular, you can compute $9^3$ and then cube that.