Prime numbers of form $a^n-1$. Let $a$ and $n$ be integers greater than $1$. Suppose that $a^n - 1$ is prime. Show that $a=2$ and $n$ is prime.

Since $a^n-1=(a-1)(a^{n-1}+\dots+1)$ if $a>2$ then $a^n-1$ is divisible with $a-1$. $a^n-1$is prime, so $a$ should be 2.
If $n=p\times q$ then $2^n-1= (2^p)^q-1=(2^p-1)((2^p)^{q-1}+\dots+1)$ and $2^n-1$ will divisible with $2^p-1$. $2^n-1$ is prime so $n$ wouldn't be written as $p\times q$, so $n$ is prime.
