If an=22n+1 then for n>1 , last digit of an is

For $n = 2, a_{n} = 2^{4} + 1 = 16 + 1 = 17$ . . Assume

That $a_{k} = 2^{2^{k}} + 1 = 10 m + 7$ where $k > 1$

and m is some positive integer.

Now, $\begin{array}{l} a_{k + 1} = 2^{2^{k + 1}} + 1 = {(2^{2^{k}})}^{2} + 1 \\ = (10 m + 6)^{2} + 1 = 10 (10 m^{2} + 12 m + 3) + 7 \end{array}$

Thus, last digit of $a_{n}$ is $7 \forall n > 1$

If $a_{n} = 2^{2^{n}} + 1$ then for $n > 1$ , last digit of $a_{n}$ is