Is the last digit of this number :$ {{4^4}^n}+1 $ always $7 $ for $n>1$ and could this be prime?

Question

Some computations in wolfram alpha for $n=2,3,4,5 ,6$ showed that the last

digit of this number $ {{4^4}^n}+1 $ for $n>1$ always $7$ .

My question here :How do I know if it's last digit always is $7$ ?

Note: My Goal is to know for which values of $n$: $ {{4^4}^n}+1 $ could be prime ?

Thank you for any help

Answer 1

The answer is yes. This follows because

• $4^r$ is always even, in particular for $r = 4^n$
• $4^r \equiv (-1)^r = 1 \mod 5$ for all even $r$, in particular for $r = 4^n$.

Therefore $4^{4^n} \equiv 6 \mod 10$ for all $n$ and hence $4^{4^n} + 1 \equiv 7 \mod 10$.

Answer 2

Yes, it is true. For $n≥1$, $4^{(4^n)} = 16^{2^{2n-1}}$ is a power of 16.

Since $16^2 = 256 \equiv 6 \pmod {10}$ and $16\cdot 6 = 96 \equiv 6 \pmod{10}$, every power of $16$ ends with $6$. Therefore $4^{(4^n)}+1$ always has $7$ as last digit.