<s>[INST] 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... 
 [/INST] $2^{4^k}+5 \equiv (-1)^{4^k}+5 \equiv 1+5 \equiv 0 \pmod 3$
Proof by induction that $2^{4^k}+5 \equiv 0 \pmod 7$ for all non negative ingeters $k$.
If $k=0$, then $2^{4^k}+5 \equiv 2+5 \equiv 0 \pmod 7$
Suppose that $2^{4^m}+5 \equiv 0$ for some non negative integer $m$.
Then 
$2^{4^{m+1}}+5 \equiv
 2^{4\cdot 4^m}+5 \equiv
 (2^4)^{4^m}+5 \equiv 
 16^{4^m}+5 \equiv
 2^{4^m}+5 \equiv 0 \pmod 7$
Hence, by mathematical induction $2^{4^k}+5 \equiv 0 \pmod 7$ for all non negative integers $k$.
It follows that $2^{4^k}+5 \equiv 0 \pmod{21}$ for all non negative integers $k$.
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