For all $m\ge 2$, the last digit of $ 2^{2^m} $ is 6 How to prove that for  $ \forall m\in \Bbb N $ such that $m\ge 2$, the last digit of $  2^{2^m} $ is 6?

$$2^{(2^m)}=(2^4)^{(2^{m-2})}=(16)^{(2^{m-2})}\equiv6^{(2^{m-2})}\pmod{10}$$
Now, $6^{n+1}-6=6(6^n-1)\equiv0\pmod {10}$ as $6^n-1$ is divisible by $(6-1)=5$ if integer $n\ge 0$
$\implies 6^r\equiv6\pmod{10}$ if $r\ge1$
So we need  $2^{m-2}\ge1\implies m-2\ge 0\implies m\ge2$
