Show that by induction method that $2^{2^n}+1$ has $7$ in unit's place for all $n\geq 2$.
I have tried to show this with the following way :
Let $f(n)=2^{2^n}+1$.
Then for $n=2,f(2)=2^{2^2}+1=17\Rightarrow f(2)\equiv 7(\mod 10) $
Suppose for $n=m$, the result is true i.e., $f(m)=2^{2^m}+1=10p+7$, where $p$ is an integer.
How can I show this result for $n=m+1$ ?