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

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 $7$.
