$2^n-1$ is prime or composite as $n$ is prime or composite [without using induction method]

Question

I have proved this for prime $n$.

I need help to prove $2^n-1$ is composite when $n$ composite

As far I can think - suppose $2^n-1$ is composite but $n$ is prime. $\\$
Therefore, $1$ divides $$2^n-1=(2-1)(2^{n-1}+2^{n-2}+....+2+1)$$ which means $2^n-1$ is divisible by $1$ and the number itself. Hence $2^{n-1}+2^{n-2}+...+2+1>1$ and it cannot be represented as the product of its factors. Hence $2^n-1$ is prime - a contradiction, hence $n$ is composite.

Am I correct here? I need help. Any help is appreciated.

Answer 1

Hint Write $n=ab$ where $1<a,b<n$ and use the factorization $$ (x^b-1)=(x-1)(x^{b-1}+x^{b-2}+\dotsb+1) $$ where $x=2^a$. Show that the factorization is indeed witness to $2^n-1$ being composite.

Answer 2

Hint: $$2^n-1=\left (2^a\right)^b-1=(2^a-1)(2^{(a-1)b}+2^{(a-2)b}+\ldots +1)$$