Prove using induction that $2^{4^n}+5$ is divisible by 21 I have to show, using induction, that $2^{4^n}+5$ is divisible by $21$. It is supposed to be a standard exercise, but no matter what I try, I get to a point where I have to use two more inductions.
For example, here is one of the things I tried:
Assuming that $21 |2^{4^k}+5$, we have to show that $21 |2^{4^{k+1}}+5$.
Now, $2^{4^{k+1}}+5=2^{4\cdot 4^k}+5=2^{4^k+3\cdot 4^k}+5=2^{4^k}2^{3\cdot 4^k}+5=2^{4^k}2^{3\cdot 4^k}+5+5\cdot 2^{3\cdot 4^k}-5\cdot 2^{3\cdot 4^k}=2^{3\cdot 4^k}(2^{4^k}+5)+5(1-2^{3\cdot 4^k})$.
At this point, the only way out (as I see it) is to prove (using another induction) that $21|5(1-2^{3\cdot 4^k})$. But when I do that, I get another term of this sort, and another induction.
I also tried proving separately that $3 |2^{4^k}+5$ and $7 |2^{4^k}+5$. The former is OK, but the latter is again a double induction.
Is there an easier way of doing this?
Thank you!
EDIT
By an "easier way" I still mean a way using induction, but only once (or at most twice). Maybe add and subtract something different than what I did?...
Just to put it all in a context: a daughter of a friend got this exercise in her very first HW assignment, after a lecture about induction which included only the most basic examples. I tried helping her, but I can't think of a solution suitable for this stage of the course. That's why I thought that there should be a trick I am missing... 


Is there an easier way of doing this?

Note that $2^{4^{k+1}}=(2^{4^k})^4$.
Inductive step : 
Supposing that $2^{4^k}+5=21m$ gives 
$$\begin{align}2^{4^{k+1}}+5&=(2^{4^k})^4+5\\&=(21m-5)^4+5\\&=\sum_{i=0}^{4}\binom{4}{i}(21m)^i(-5)^{4-i}+5\\&\equiv (-5)^{4}+5\quad\pmod{21}\\&\equiv 0\pmod{21}\end{align}$$
Added :
If you want to use neither the binomial theorem nor mod, then supposing that $2^{4^k}+5=21m$ gives
$$\begin{align}2^{4^{k+1}}+5&=(2^{4^k})^4+5\\&=(21m-5)^4+5\\&=21(9261m^4-8820m^3+3150m^2-500m+30)\end{align}$$
So, $2^{4^{k+1}}+5$ is divisible by $21$.
