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 $2^{2^{n+1}}=2^{2^n}2^2.$
 A: Continuing from where you left, we just need to prove that $F_{n+1}$ holds true.
$$F_{n+1} = 2^{2^{n+1}}+1$$
$$F_{n+1} = 2^{2^n.2}+1$$
$$F_{n+1} = (2^{2^n})^2 + 1$$
$$F_{n+1} = (2^{2^n} + 1)^2 - 2.2^{2^n}$$
First term on the right hand side is $F_n^2$ which has a last digit of $7^2=9$.
Second term on the right hand side is $2(F_n-1)$, which has a last digit of $2(7-1)=2$.
Hence on a whole the last digit of right hand side, i.e., $F_{n+1}$ is :
$$9-2 = 7$$
Hope the answer is clear !
NOTE: You can cut short the above solution at Step:$3$ by saying that $(F_n-1)^2$ has a last digit of $6$ and hence $(F_n-1)^2+1 = (2^{2^n})^2 + 1$ has a last digit of $7$.
A: $\!\bmod 10\!:\ F_N\equiv 7\,\overset{\large \rm subtract\ 1}\Longrightarrow\,\color{#c00}{2^{\large 2^{\Large N}}\!\!\!\equiv 6}\,\overset{\large\rm square}\Longrightarrow\,2^{\large 2^{\Large N+1}}\!\!\!=(\color{#c00}{2^{\large 2^{\Large N}}})^{\large 2}\equiv \color{#c00}6^{\large 2}\!\equiv 6\,\overset{\large\rm add\ 1\!}\Longrightarrow\,F_{N+1}\equiv7 $ 
A: $$ (2^{2^n}+1)^2 =  2^{2^{n+1}} + 2\cdot 2^{2^n}+1$$
Modulo 10 the left-hand side is 9 by the inductive hypothesis. Also
$$2\cdot 2^{2^n} = 2(2^{2^n}+1)-2$$
Modulo 10, this is 2.
What we have is:
$$(2^{2^n}+1)^2 - 2\cdot 2^{2^n}\,\,\, (\text{mod 10}) =  9-2=7$$
But that's the same as
$$2^{2^{n+1}} + 1 \,\,\, (\text{mod 10})$$
A: 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 $\displaystyle 2^{n-2}\ge1$
