How to show that there is no odd $n$ bigger than 100 such that the group $U(n)$ will contain $2^n$ number of elements? Consider $U(n):=\{1\leq r\leq n: (r, n)=1\}$. Under multiplication modulo $n$ it forms an abelian group. Its order will be $\varphi(n)$ i.e. $\varphi(n)=|U(n)|$. 
Assume that $n\geq 100$. Then my question is : does there exists any odd natural number $m_o$ such that the group $U(n)$ will have the order $2^n$ ? The reason I have  chosen $n\geq 100$ is that : I have already checked the result below 100 and the answer I have obtained so far is NO if $33\leq n\leq 100$. No need to care about $n\leq 33$. I think the result is still NO if we go beyond 100.
But the question is : how can I show no such odd $m_o$ would exists ? 
What I have approached is the following:
Since $m_o$ is odd so we can write $$m_o=\prod\limits_{i=1}^r p_i^{a_i}$$ where $p_i$ are odd primes and $a_i$ are positive integers. Then we must have 
$$U(m_o)\simeq U(p_1^{a_1})\times \cdots \times U(p_r^{a_r})$$ where $\times$ denotes the external direct product of groups. I don't know if we can create some contradiction over here.
Also note that , the question is equivalent to say that if $m_o$ is one such odd number then there will be $2^n$ number of positive integers less than $m_o$ and relatively prime to $n$. We have to show no such $m_o$ is possible to find after 100. 
in this way also no contradiction i am able to get. 
Please help me. thanks in advance.

You want positive integers $n$ such that $\varphi(n)$ is a power of $2$. But if $n=p_1^{a_1}\cdots p_m^{a_m}$, then $\varphi(n)=\displaystyle\prod_{i=1}^m (p_i^{a_i}-p_1^{a_1-1})$. 
Therefore, the only primes you can have are $2$ and those such that $p_i-1$ is a power of $2$. The only such primes which are known are $3, 5, 17, 257, 65537$.
