USAMO

The assertion is false, and the smallest $n$ for which it fails is $n = 25$. Given $n \ge 2$, let $r$ be the remainder when $2^n$ is divided by $n$. Then $2^n = kn + r$ where $k$ is a positive integer and $0 \le r < n$. It follows that
$$2^{2n} = 2^{kn+r} \equiv 2^r \pmod{2^n - 1},$$
where $2^r < 2^{2n} - 1$. Thus, the remainder when $2^{2n}$ is divided by $2^n - 1$ is $2^r$. Now, $2^r$ is a power of 4 if and only if $r$ is even, so to disprove the assertion it is enough to find an $n$ for which the corresponding $r$ is odd. We describe now a method to find such an $n$.
If $n$ is even then so is $r = 2^n - kn$. If $n$ is an odd prime then $2^n \equiv 2 \pmod n$ by Fermat's Little Theorem; hence $r \equiv 2^n \equiv 2 \pmod n$ and $r = 2$. Therefore, we may rule out these values of $n$.
It remains to try cases in which $n$ is odd and composite. In the first three instances $n = 9, 15, 21$ there is no contradiction to the assertion:
$$n = 9: \quad 2^6 \equiv 1 \pmod 9 \quad \Rightarrow \quad 2^9 \equiv 2^6 \cdot 2^3 \equiv 8 \pmod 9$$
$$n = 15: \quad 2^4 \equiv 1 \pmod{15} \quad \Rightarrow \quad 2^{15} \equiv (2^4)^3 \cdot 2^3 \equiv 8 \pmod{15}$$
$$n = 21: \quad 2^6 \equiv 1 \pmod{21} \quad \Rightarrow \quad 2^{21} \equiv (2^6)^3 \cdot 2^3 \equiv 8 \pmod{21}$$
However, for $n = 25$, we see that
$$2^{10} = 1024 \equiv -1 \Rightarrow 2^{20} \equiv 1 \Rightarrow 2^{25} \equiv 2^5 \equiv 7 \pmod{25},$$
so 7 is the remainder when $2^{25}$ is divided by 25 and $2^7$ is the remainder when $2^{2^{25}}$ is divided by $2^{25} - 1$. This gives the desired counterexample.