<s>[INST] Is there a quicker proof to show that $2^{10^k} \equiv 7 \pmod{9}$ for all positive integers $k$? I noticed this pattern while playing with digit sums and noticed that the recursive digit sums (until you arrive at a single digit number) of numbers like, $2^{10}$, $2^{100}$, $2^{1000}$ and so on is always $7$. So, I decided to find out if it is true that for all positive integers $k$,
$$ 2^{10^k} \equiv 7 \pmod{9} $$
My proof is as follows:

Lemma 1. $\, 10^k \equiv 4 \pmod{6}$ for all integers $k \geq 1$
Proof. For all integers $k \geq 1$, the number $10^k + 2$ must be divisible by $6$ since it is even (implying divisibility by $2$), and its digit sum is $3$ (implying divisibility by $3$). Therefore, you can show that
\begin{align*}
10^k + 2 &\equiv 0 \pmod{6} \\
10^k &\equiv 4 \pmod{6} \,\, \text{    for all integers  } k \geq 1
\end{align*}
Lemma 2. $\, 2^{4 + 6k} \equiv 7 \pmod{9}$, for all integers $k \geq 0$
Proof. If $a \equiv c \pmod{n}$ and $b \equiv d \pmod{n}$, then $a\,b \equiv c\,d \pmod{n}$.
And, by extension, $a\,b^k \equiv c\,d^k \pmod{n}$ (for integers $k \geq 0$). Therefore, with
\begin{align*}
2^4 \equiv 16 \equiv 7 \pmod{9}
\end{align*}
and
\begin{align*}
2^6 \equiv 64 \equiv 1 \pmod{9}
\end{align*}
we can show that,
\begin{align*}
2^{4 + 6k} \equiv 7 \cdot 1^k \equiv 7 \pmod{9} \,\, \text{    for all integers  } k \geq 0
\end{align*}
Theorem. $\, 2^{10^k} \equiv 7 \pmod{9} $ for all integers $k \geq 1$
Lemma 1 implies that for all integers $k \geq 1$, $10^k = 4 + 6n$ where $n$ is some positive integer. Therefore,
\begin{align*}
2^{10^k} &= 2^{4 + 6n} \\
2^{10^k} \!\!\!\! \mod a &= 2^{4 + 6n} \!\!\!\! \mod a
\end{align*}
for any positive integer $a$. Using this result along with Lemma 2, 
\begin{align*}
2^{10^k} \equiv 7 \pmod{9} \,\, \text{    for all integers  } k \geq 1
\end{align*}
$$\tag*{$\blacksquare$}$$

I think the proof is correct, but I am not a fan of it. Mainly because, I think proving Lemma 2 is a much too long a way to prove this theorem - since it proves a generalization of the theorem first. Also, I discovered that lemma numerically, which feels a bit like cheating.
Either way, is there a quicker, more elegant proof which does not:


*

*Use Lemma 2, or prove some generalization of the theorem first.

*Uses Euler's Theorem. I feel that using Euler's Theorem is using a needlessly complicated theorem to prove something as simple as this.
I am sure some of the people here can come up with single line proofs. I am curious to see if there's such a proof.
 [/INST] Induction? Base case when $k=1$ is clear, for inductive step we have: $2^{10^k}=(2^{10^{k-1}})^{10}\equiv 7^{10}\equiv 7\bmod 9$
</s>