If $n$ is an even natural number then $2^{n-1}$ is divisible for $2^{\varphi(n)}$?

Question

If $n$ is an even natural number ($n \neq 0$) then is $2^{n-1}$ divisible by $2^{\varphi (n)}$?

Since $n$ is an even number then $n-1$ is an odd number. Hence, $(2,n)=1$. By Euler's theorem, $2^{\varphi (n-1)} - 1$ is divisible for $n-1$.

But I don't know what I should do. Any help would be greatly appreciated.

Answer 1

Since $n>1$, $\varphi(n)\leqslant n-1$, and therefore $2^{\varphi(n)}\mid2^{n-1}$.