I have some problems understanding the following proof:
Definition: A composite number $n \in \mathbb{N}$ is called pseudo prime if $n \mid 2^{n-1}-1$ holds.
Theorem: If n is a odd pseudo prime number, then $2^n-1$ is also an odd pseudo prime number, too.
Proof: Let n be an odd pseudo prime number. Then we get $2^{n-1}-1=kn$ with $n \in \mathbb{N}$. It follows $2^{2^n-2}=2^{2kn}$ and further $2^{2^n-2}-1=(2^n)^{2k}-1$. We get $2^n-1 \mid 2^{2^n-2}-1$, but also $2^n-1 \mid 2^{2^n-1}-2$. Let $d \mid n$ with $1
Now my questions: What is the reason, that the author made this step: $2^{2^n-2}=2^{2kn}$, just to get rid of the "k"? The next step is also not clear: $2^{2^n-2}-1=(2^n)^{2k}-1$.
If anyone could explain the steps to me, I would be thankful. Greetings