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

By Fermat's little theorem, we have
$$2^{18} \equiv 1 \pmod{19} \implies 2^{36} = (2^{4})^{9} \equiv 1 \pmod{19} \tag{1}\label{eq1A}$$
Since $2^6 = 64 \equiv 1 \pmod{9}$, we thus get $2^{6k} = (2^6)^k \equiv 1 \pmod{9}$, so there's an integer $j$ where $2^{6k} = 9j + 1$. Thus,
$$(2^4)^{2^{6k}} = (2^4)^{9j + 1} = ((2^4)^{9})^{j}(2^4) \equiv 2^4 \pmod{19} \tag{2}\label{eq2A}$$
