Let $\varphi$ denote Euler's totient function. It's widely believed that $2^n1$ is prime for infinitely many (prime) $n$, which, in turn, implies $$\limsup_{n \to \infty} \frac{\varphi(2^n1)}{2^n1} = 1. $$ But what about an unconditional proof? This is sort of a followup of a question asked by @kodlu yesterday.
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Let $n$ be a large prime. Then $2^n1=p_1\dots p_k$ (possibly with equal factors), where $n\mid p_i1$ for all $i$. Therefore, $k\leq \log_n(2^n1)<\frac{n}{\log_2n}$. Thus, $$ \frac{\varphi(2^n1)}{2^n1}\geq\prod_{i=1}^k\left(1\frac1{p_i}\right) \geq \left(1\frac1n\right)^{n/\log_2n}\to1, \quad n\to\infty, $$ as required.

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