The result holds for any prime $p$ that is not a Fermat prime. Take $a_0=g$ for some primitive root $g$ mod $p$. Note that $a_n=g^{2^{\frac{n(n+1)}{2}}}$. Hence, it suffices to show that $2^{\frac{n(n+1)}{2}}$ does not eventually become constant mod $p-1$. Since $p$ is not a Fermat prime, $p-1$ has some odd prime factor $q$. Then it suffices to show that $f(n)=\frac{n(n+1)}{2}$ is not eventually constant mod $q$, or equivalently constant mod $q$. This is clearly true, because $f(0)=0$ and $f(1)=1$.