prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$ number theory prove for all $k \in \mathbb{Z_{\geq 0}}$ , $2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$
attempt:
$$2^{2^{6k}\cdot4}\equiv 2^4\pmod{19}$$
we can rewrite the equation
$$(2^4)^{2^{6k}}\equiv 2^4\pmod{19}$$
How I can continue from there to apply the Euler's theorem

Induction:
\begin{align*}2^{2^{6(k+1)}\cdot 4}=2^{2^{6k+6}\cdot 4}=\left(2^{2^{6k}\cdot 4}\right)^{64}&\equiv (2^4) ^{64}\mod 19\\&\equiv 2^{256}\mod19\end{align*}
By Fermat's little theorem, $2^{18}\equiv 1\mod 19$. Since $256=18*14+4$, then
\begin{align*}2^{2^{6(k+1)}\cdot 4}&\equiv (2^{18})^{14}2^4\mod19\\&\equiv 1^{14}2^4\mod 19\\&\equiv 2^4\mod19\end{align*}
