Proving $2^{2^n} \equiv 1(\mod 5)$

Question

Prove that $2^{2^n} \equiv 1(\mod 5)$, when $n>1$

Any help would be appreciated.

Answer 1

$$2^2=4\equiv-1\pmod5\implies (2^2)^m\equiv\begin{cases}1 &\mbox{if } m\text{ is even} \\ -1 & \mbox{if } m \text{ is odd} \end{cases} $$

So here, we need $2^n$ to be divisible by $2\cdot2=2^2$ which is true if $?$

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Alternatively, $\displaystyle 2^{2^n}=(2^2)^{2^{n-1}}\equiv(-1)^{2^{n-1}}\pmod5$

So, we need $\displaystyle2^{n-1}$ to be even $\iff$ integer $n-1>0$