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 $2^{2^{n+1}}=2^{2^n}2^2.$

$$ (2^{2^n}+1)^2 =  2^{2^{n+1}} + 2\cdot 2^{2^n}+1$$
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
$$2^{2^{n+1}} + 1 \,\,\, (\text{mod 10})$$
