Number Theory: Show that $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ Show that for all $n\in\mathbb{N}$, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$.
I think I've proved this problem, but I was unsure if my proof was correct:
Proof
Let $n=1$. Then, $10^3=1000\equiv1\pmod{3^3}$ since $3^2=9\mid999$ and $3\mid111\implies 3^3\mid999$; but, $3^4\not\mid999$.
Now, assume $10^{3^n}\equiv1\pmod{3^{n+2}},3^{n+3}\not\mid10^{3^n}-1$.
We want to prove that this is true for $n+1$:
$10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$
$10^{3^{n+1}}-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n})^3-1\equiv0\pmod{3^{n+3}}$
$(10^{3^n}-1)((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod{3^{n+3}}$.
By our assumption, $3^{n+2}\mid(10^{3^n}-1)$ but $3^{n+3}\not\mid10^{3^n}-1$, so $\gcd(3^{n+3},10^{3^n}-1)=3^{n+2}$.
So,
$((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$.
But $((10^{3^n})^2+10^{3^n}+1)$ is of the form $100\dots00100\dots001$, so the sum of its digits is $3$, so it can be divided by $3\implies ((10^{3^n})^2+10^{3^n}+1)\equiv0\pmod3$. (Just added this:) However, it cannot be divided by $3^2=9$ since the sum of these digits is not divisible by $9$.
Thus, $10^{3^{n+1}}\equiv1\pmod{3^{n+3}}$ and $3^{n+4}\not\mid 10^{3^{n+1}}-1$.
So, by induction, $10^{3^n}\equiv 1\pmod{3^{n+2}}$ but $3^{n+3}\not\mid 10^{3^n}-1$ for all $n\in\mathbb{N}$.

The "divides" part is fine. You have enough machinery set up to prove that $3^{n+3}$ does not divide $10^{3^n}-1$, but you have not proved it explicitly. Hint: You showed that a certain expression is divisible by $3$. Show it is not divisible by any higher power of $3$.
