Prove by induction that $2^{4^n}+5$ is divisible by 21 .

Question

After I did the first two steps in the induction , I am stuck in the last step ; to prove for $n+1$ that $2^{4^n}+5$ is divisible by $21$ , so
I know that $2^{4^n}+5$ is divisible by $21$ is true .
I want to prove for $n+1$ ( $n$ is natural ):
$$2^{4^{n+1}}+5=$$ $$=2^{4^{n}\cdot 4}+5=$$ $$=2^{4^{n}\cdot 4}+5=$$ $$=2^{{4}^{{4}^{n}}}+5=$$ $$=16^{{4}^{n}}+5.$$
this is the last step I reached , I would be happy to get some recommendation on how to continue from here .

Answer 1

Hint:

The statement is equivalent to $2^{4^n} = -5 + 21k$ for some $k\in\mathbb Z$. Then raise both sides to the $4$'th power. (Looks cleaner with modular arithmetic if you know it).