<s>[INST] 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?
 [/INST] $\bmod 4\!:\ 9^{56}\equiv 1^{56}\equiv 1\,\Rightarrow\, 9^{56}=4m+1$.   
$$\bmod 25\!:\ 9^{56}\stackrel{(1)}\equiv 9^{56\pmod{\phi(25)}}\equiv 9^{56\pmod {20}}\equiv 9^{-4}\equiv \left(\frac{1}{81}\right)^2$$   
$$\equiv \left(\frac{1+5\cdot 25}{6+3\cdot 25}\right)^2\equiv \left(\frac{126}{6}\right)^2\equiv 21^2\equiv (-4)^2\equiv 16\equiv 4m+1$$   
$$\iff 4m\equiv 15\equiv 40\iff m\equiv 10\iff m=25l+10$$    
$9^{56}=4(25l+10)+1=100l+41$. In $(1)$ I used Euler's theorem.
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