<s>[INST] Proving a property of about the Fermat numbers 
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.  

For our base step we let $n=2$.  Now we have $2^{2^2}=16$.  So the assertion holds for our base case.  Then we assume it holds for $n$.  For the $n+1$ case, is there a way to demonstrate this without resorting to this:  $$(2^{2^n})^{{2^{n+1}}-2^n}.$$
That is the solution available in the back of the book.  My attempt was to look at $2^{2^{n+1}}=2^{2^n}2^2.$
 [/INST] Observe that $\displaystyle2^1\equiv2,2^2\equiv4,2^3\equiv8,2^4=16\equiv6,2^5=32\equiv2\pmod{10}$
So, $\displaystyle2^m\equiv6\pmod{10}\iff 4|m$ 
$\displaystyle\implies2^{2^n}\equiv6\pmod{10}\iff 4|2^n\iff n\ge2$

Alternatively, as $\displaystyle6^{m+1}-6=6(6^m-1)\equiv0\pmod{30}$ as $6^m-1$ is divisible by $6-1=5$
$\displaystyle\implies 6^{m+1}\equiv6\pmod{30}\equiv6\pmod{10}$ for integer $m\ge0$
Again, $\displaystyle2^4=16\equiv6\pmod{10}$
So, $\displaystyle2^{2^n}=(2^4)^{2^{n-2}}=16^{(2^{n-2})}\equiv6^{(2^{n-2})}\pmod{10}$ for $\displaystyle 2^{n-2}\ge1\iff n\ge2$
Again, $\displaystyle6^{(2^{n-2})}\equiv6\pmod{10}$ for $\displaystyle 2^{n-2}\ge1$
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