Divisibility of $3^n-1$ by $2^n$. It is a curious fact that $3^n-1$ is divisible by $2^n$ iff $n=1,2,4$. (It turns out to have applications in algebraic topology, differential topology and algebra.) Does anyone know of a very short proof of this fact? Methods should come from (elementary) number theory or abstract algebra.

For $n\geq 3$ odd, it is obvious, since then $3^n\equiv 3\pmod 8$.
Assume $2^{2n}\mid 3^{2n}-1$. Then since $3^n+1\equiv 2,4\pmod {8}$, we know that at most $2^2\mid 3^n+1$, so $2^{2n-2}\mid 3^{n}-1$. For $n\geq 2$, $2n-2\geq n$, so: $2^{n}\mid 3^n-1$.
So we have first that for $n$ divisible by any odd prime, this is not true.
All that remains are powers of $2$. We just need to check $n=8$. And:
$$3^8-1 = (3^4-1)(3^4+1)=(16\cdot 5)(2\cdot 41)= 2^5 \cdot 5\cdot 41$$
