The residue of $9^{56}\pmod{100}$ How can I complete the following problem using modular arithmetic?

Find the last two digits of $9^{56}$. 

I get to the point where I have $729^{18} \times 9^2 \pmod{100}$.  What should I do from here?

By Carmichael's function, the order of any residue of $100$ divides, and is a maximum of, $20$ (the same as for $25$). We see that $9$ is a square, so the maximum order of $9 \bmod 100$ is $10$ (or divides $10$ if it is less). This gives
$$9^{56}\equiv 9^6 \bmod 100$$
Then we can simply multiply a few small powers:
$$9^2 \equiv 81,\quad 9^4 \equiv 61, \quad 9^6 \equiv  41 \equiv 9^{56}$$
(This result also shows - since $9^6 \not\equiv 9$ - that the order of $9 \bmod 100$, not being $2$ or $5$, must be $10$, as others have shown by direct calculation)
