number theory - How to get the result using Pepin's test

How do I achieve this result via Pepin's test by using Euler's Theorem and such to simplify $3^{2^{31}}$ and get the desired congruence of 10,324,303?

$3^{(F_5-1)/2} = 3^{2^{31}} = 3^{2,146,483,648} \equiv 10,324,303 ≢ -1 (mod 4,294,967,297)$

Where $F_5$ is Fermat's 6th number $2^{32}+1$

Using powermod functions gives the desired result for $a^{b} mod(m)$ but I'd like to do this one by hand to better understand how powermod simplifies.

Thanks!