What is the remainder when $ 6^{65} $ is divided by 80? In the solutions it used Chinese remainder theorem in the following way: $6^{65}\equiv 1 \pmod 5 $ and $ 6^{65} \equiv 0 \pmod{16} $. I can see how they got the first congruence but am really struggling to see it for the second one. Any help would be much appreciated!

Note:
$$\begin{align} 6 &\equiv 1\pmod 5 \\
6^{61} &\equiv 1 \pmod 5 \\
6^{61}\cdot 2^4 &\equiv 1\cdot 2^4 \pmod {5\cdot 2^4}\\
3^4 &\equiv 1\pmod{80}\\
6^{61}\cdot 2^4\cdot 3^4 &\equiv 1\cdot 2^4\cdot 1 \pmod{80}\\
6^{65} &\equiv 16 \pmod{80}.\end{align}$$
