A problem by AoPS Master

By binomial expansion we get: $(13-1)^{144}$ simplifies to $13^1\cdot(-1)^{143}+13^0\cdot(-1)^{144}$ when we cancel the terms that have at least two 13s. This simplifies to $-12$ which is $157$ modulo $169$ .

SOLUTION:

If we see then we can find $\displaystyle 12^{156} \equiv 1 \pmod {169 }$ .

Or we can say $\displaystyle 12^{144} \times 12^{12} \equiv 1 \pmod {169 } \implies 12^{144} \times 14 \equiv 1 \pmod {169 }$ .

Multiply $12$ both sides as $12 \times 14 = 168$ , $\displaystyle 12^{144} \times 14 \times 12 \equiv 12 \pmod{169 } \implies 12^{144} \equiv -12 \pmod {169 }\implies 12^{144} \equiv 157 \pmod{169 }$ .