<s>[INST] If $n$ is an odd pseudoprime , then $2^n-1$ is also odd pseudoprime 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 $1<d<n$, then it follows $2^d-1\mid 2^n-1$ and $1<2^d-1<2^n-1$. So $2^n-1$ is also an pseudo prime number.
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
 [/INST] This solution is essentially the same as the one quoted by OP, with some detail added and some removed. The most useful change is the introduction of $N$, which allows us to have one less level of exponentiation. 
We have $2^{n-1}\equiv 1 \pmod{n}$. Let $N=2^n-1$. We first show that $N$ is composite. Since $n$ is composite, $n=ab$ for some $a>1$, $b>1$. Then $2^a-1$ is a non-trivial divisor of $2^n-1$.
Now we need to show that $2^{N-1} \equiv 1 \pmod{N}$. Since $n$ divides $2^{n-1}-1$, set $2^{n-1}-1=nk$. Multiply by $2$. We get $2^n-2=2nk$. But $2^n-2=N-1$, so $N-1=2nk$.  
Thus $2^{N-1}=(2^n)^{2k}=(1+N)^{2k}\equiv 1 \pmod{N}$, which is the desired result.
Comment: One reason we let $N=2^n-1$ is that then the messy $2^{2^n-2}$ of the post becomes the understandable $2^{N-1}$, which is the object we need to compute modulo $N$  to show that $2^n-1$, that is, $N$, is a pseudo-prime to the base $2$.
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