Let $p \geq 2$. Prove that if $2^p-1$ is prime then $p$ must also be prime. Would the following be a valid proof?
Let $r$ and $s$ be positive integers, then the polynomial $x^{rs}-1=(x^r -1)(x^{s(r-1)}+x^{s(r-2)}+\cdots+x^r+1)$.
So if $p$ is composite (say $rs$ with $1<s<p$), then $2^p-1$ is also composite (because it is divisible by $2^s-1$)

By contrapositive, if $p$ is not prime, then $2^p-1$ is not prime. Since $p$ is not prime, is composite, so $p=nr$ with $1<r<p, 1<n<p$. Now $2^p-1=(2^r)^n-1$, use the change $x=2^r$ we get $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1).$$
Clearly $(x^{n-1}+x^{n-2}+x^{n-3}+ \cdots+x+1)>1.$ Since $x=2^r \wedge r>1, 2^r>1 \Rightarrow x-1>1.$ Thus, $x^n-1=2^{rn}-1=2^p-1$ is the product of terms greater than 1. Finally, $2^p-1$ is not prime
