To show $n\in\mathbb{N}\setminus \{6\}\Rightarrow 2^{\varphi(n)}\ge n$
I can't follow the proof from http://mathematicalspectacles.blogspot.de/2012/05/interesting-study-of-zsigmondy-primes.html
Lemma 2. There is a part with
If $k=1$ in other word $2^{\varphi(n)}=n+1$. The $\varphi(n)$ numbers $2^i$ for $0\le i \le \varphi(n)-1$ are coprime with $n=2^{\varphi(n)}−1$ and are between 1 and n.
This is clear.
WHY IS : $1\le 2^{\varphi(n)}−2\le n $ The $\le n$ is clear. But why $1\le 2^{\varphi(n)}$?
"and $\gcd(2^{\varphi(n)}−2,n)=1$, we have $2^{\varphi(n)}−2=2^i$ for some $0\le i\le \varphi(n)−1$." This is also clear.
"Hence $2^{\varphi(n)-1}−1=2^{i−1}$, so $2^{\varphi(n)-1}−1=1$ [and] $n=3$."
MY QUESTIONS IS WHY: "It follows that $2^{\varphi(n)}>3n$ holds for odd $n>3$."
It would be great If someone can explain me the 2 points which I can't follow.