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Using Fermat theorem, $a^p -a = 0 (\mod p) $ and if a and p are co-prime, then $ a^{pāˆ’1} āˆ’ 1 =1(\mod p) $ So if u choose n to be a prime number(say p), then $a^{b^c} \mod p = a^{ (b^{c} \mod (p-1))} \mod p $ , Then you can use Fast Exponentiation trick in two levels, once for $b^c \mod p-1 $ then for $a^{b^c} \mod p $ I also suggest you to look at cses ...

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