We can now look at an example, let's say \(a=3\) and \(p=7\), using Fermat's Theorem we should get that \(3^{7-1}\equiv1\) (mod \(7\)) which then simplifies to \(3^6\equiv1\) (mod \(7\)). When computing this we will end up with \(\frac{729}{7}\), which does come out to have a remainder of \(1\).