What is the remainder of $3\text{^}(15\text{^}2019)$ when divided by $29$?

Question

I think you are supposed to use Fermat's little theorem but I don't know how to apply it. I tried using exponent laws but I don't know about any such laws that work in this scenario.

Answer 1

$$F=3^{15^{2n+1}}\equiv3^{15^{2n+1}\pmod{28}}\pmod{29}$$

Now $15^2\equiv1\pmod{28}$

$$\implies F\equiv3^{15(1)^n}\pmod{29}$$

Now $3^3\equiv-2\pmod{29}$

$\implies3^{15}=(3^3)^5\equiv(-2)^5\equiv-3\equiv-3+29$